Let us first split the middle term - 5x as -2x -3x [because (-2x) x (-3x) = 6x² = (2x² ) x 3].
So, 2x² - 5x + 3 = 2x² - 2x - 3x + 3 = 2x (x - 1) -3(x - 1) = (2x - 3)(x - 1)
Now, 2x² - 5x + 3 = 0 can be rewritten as (2x - 3)(x - 1) = 0.
So, the values of x for which 2x² - 5x + 3 = 0 are the same for which (2x - 3)(x - 1) = 0,
i.e., either 2x - 3 = 0 or x - 1 = 0.
Now, 2x - 3 = 0 gives x = \(\frac{3}{2}\) and x - 1 = 0 gives x = 1.
So, x = \(\frac{3}{2}\) and x = 1 are the solutions of the equation.
In other words, 1 and \(\frac{3}{2}\) are the roots of the equation 2x² - 5x + 3 = 0.
Verify that these are the roots of the given equation.
Note that we have found the roots of 2x² - 5x + 3 = 0 by factorising 2x² - 5x + 3 into two linear factors and equating each factor to zero.

kx(x - 2) + 6 = 0
or kx² - 2kx + 6 = 0
Comparing this equation with ax² + bx + c = 0, we get
a = k, b = - 2k and c = 6
= ( - 2k)² - 4 (k) (6)
= 4k² - 24k
For equal roots,
b² - 4ac = 0
4k² - 24k = 0
4k (k - 6) = 0
Either 4k = 0 or
k = 6 = 0
k = 0 or k = 6
However, if k = 0, then the equation will not have the terms 'x²' and 'x'.
Therefore, if this equation has two equal roots, k should be 6 only.

2x² + kx + 3 = 0
Comparing equation with ax² + bx + c = 0, we get
a = 2, b = k and c = 3
Discriminant = b² - 4ac = (k)² - 4(2) (3)
= k² - 24
For equal roots,
Discriminant = 0
k² - 24 = 0
k² = 24
\(k=\pm \sqrt{24}=\pm 2 \sqrt{6}\)

Given equations is \(x^{ 2 }-3x-10=0\)
\(\Rightarrow x^{ 2 }-5x+2x-10=0\)   
\(\left[ \because -5x(2)=-10\\ and-5+2=-3 \right] \)
\(\Rightarrow x(x-5)+2(x-5)=0\)
\(\Rightarrow x-5=0\quad \Rightarrow x=5\)
\(\Rightarrow x+2=0\Rightarrow x=-2\)
Hence, the roots of the equation 
\(x^{ 2 }-3x-10=0\quad are-2\quad and\quad 5\)

(i) LHS = (x – 2)² + 1 = x² – 4x + 4 + 1 = x2 – 4x + 5
Therefore, (x – 2)² + 1 = 2x – 3 can be rewritten as
x² – 4x + 5 = 2x – 3
i.e., x² – 6x + 8 = 0
It is of the form ax² + bx + c = 0.
Therefore, the given equation is a quadratic equation
(ii) Since x(x + 1) + 8 = x² + x + 8 and (x + 2)(x – 2) = x² – 4
Therefore, x² + x + 8 = x² – 4
i.e., x + 12 = 0
It is not of the form ax² + bx + c = 0.
Therefore, the given equation is not a quadratic equation
(iii) Here, LHS = x (2x + 3) = 2x² + 3x
So, x (2x + 3) = x² + 1 can be rewritten as
2x² + 3x = x² + 1
Therefore, we get x² + 3x – 1 = 0
It is of the form ax² + bx + c = 0.
So, the given equation is a quadratic equation
(iv) Here, LHS = (x + 2)³ = x3 + 6x² + 12x + 8
Therefore, (x + 2)³ = x³ – 4 can be rewritten as
x³ + 6x² + 12x + 8 = x3 – 4
i.e., 6x² + 12x + 12 = 0 or, x² + 2x + 2 = 0
It is of the form ax² + bx + c = 0.
So, the given equation is a quadratic equation

Here a = 3, b = - 2 and \(c=\frac{1}{3} \text { . }\)
Therefore, discriminant b² - 4ac = (-2)² - 4 x 3 x \(\frac{1}{3}\) = 4 - 4 = 0.
Hence, the given quadratic equation has two equal real roots.
The roots are \(\frac{-b}{2 a}, \frac{-b}{2 a}, \text { i.e., } \frac{2}{6}, \frac{2}{6}, \text { i.e., } \frac{1}{3}, \frac{1}{3} .\)

The given equation is of the form ax² + bx + c = 0, where a = 2, b = - 4 and c = 3. Therefore, the discriminant
b² - 4ac = (-4)² - (4 x 2 x 3) = 16 - 24 = -8 < 0
So, the given equation has no real roots.

(x+1)²=2(x-3)
x²+2x+1=2x-6
x²+2x-2x+1+6=0
x²+7=0
Which is of the form ax²+bx+c=0. Hence the given equation is a quadratic equation.

2x² - 6x + 3 = 0
Comparing this equation with ax² + bx + c = 0, we get
a = 2, b = -6, c = 3
Discriminant = b2 - 4ac
= (-6)² - 4 (2) (3)
= 36 - 24 = 12
As b2 - 4ac > 0,
Therefore, distinct real roots exist for this equation
\( x=\frac{-b \pm b^{2}-4 a c}{2 a} \)
\(=\frac{-(-6) \pm \sqrt{(-6)^{2}-4(2)(3)}}{2(2)} \)
\(=\frac{6 \pm \sqrt{12}}{4}=\frac{6 \pm 2 \sqrt{3}}{4} \)
\(=\frac{3 \pm \sqrt{3}}{2} \)
Therefore the root are \(=\frac{3 \pm \sqrt{3}}{2} \)

\(3 x^{2}-4 \sqrt{3} x+4=0\)
Comparing it with ax² + bx + c = 0, we get
a = 3, b = \(-4 \sqrt{3}\)  and c = 4
Discriminant = b² - 4ac
\(=(-4 \sqrt{3})^{2}-4(3)(4)\)
= 48 - 48 = 0
As b² - 4ac = 0,
Therefore, real roots exist for the given equation and they are equal to each other.
And the roots will be \(\frac{-b}{2 a}\)
Therefore, the roots are \(\frac{2}{\sqrt{3}} \text { and } \frac{2}{\sqrt{3}}\)

\(2x^2-x+{1\over 8}=0\)
\({16x^2-8x+1\over 8}=0\)
16x²-4x-4x+1=0
4x(4x+1)-1(4x-1)=0
(4x-1)(4x-1)=0
4x-1=0 or  4x-1=0
\(x={1\over 4}\ or \ x={1\over 4}\)
Hence the roots are, \({1\over 4},{1\over 4}\)

100x²-20x+1=0
100x2-10x-10x+1=0
10x(10x-1)-1(10x-1)=0
(10x-1)(10x-1)=0
10x-1=0 or 10x-1=0
\(x={1\over 10}\) or \(x={1\over 10}\)
Hence the roots are \({1\over 10},{1\over 10}\)

Given, equation is \( \sqrt2x^2+7x+5\sqrt2=0\) 
\(\Rightarrow\) \( \sqrt2x^2+5x+2x+5\sqrt2=0\) [\(\because\) 5 \(\times\)2 = 10 and 5 + 2 = 7]
\(\Rightarrow\)\(x(\sqrt2x+5)+\sqrt2(\sqrt2x+5)=0\)
\(\Rightarrow\) (\(\sqrt{2}\)x + 5) (x + \(\sqrt{2}\)) = 0
\(\Rightarrow\) \(\sqrt{2}\)x+5 = 0 or x + \(\sqrt{2}\) = 0
\(\Rightarrow\) x=\(\frac{-5}{\sqrt{2}}\) or x = -\(\sqrt{2}\)
Hence, the roots of the equation
\(\sqrt{2} x^2+7 x+5 \sqrt{2}=0 \text { are } \frac{-5}{\sqrt{2}} \text { and }-\sqrt{2} \text {. }\)

Given equation is 2x² + x - 6 = 0
\(\Rightarrow\) 2x² + 4x - 3x - 6 = 0
[\(\because\) 4 \(\times\)(-3) = -12 and 4 - 3 = 1]
\(\Rightarrow\) 2x (x + 2) -3 (x + 2) = 0
\(\Rightarrow\) (x + 2)(2x - 3) = 0
\(\Rightarrow\) x + 2 = 0 or 2x - 3 = 0
\(\Rightarrow\)   x = -2  or \(x=\frac{3}{2}\)
Hence, the roots of the equation
2x² + x - 6 = 0 are -2 and \(\frac{3}{2}\)

Given, equation is x³-4x²-x+1=(x-2)³ 
\(\Rightarrow\) x³ - 4x² - x + 1 = x³ - 3(x²)2 +3x(2)² - (2)³
[\(\because\)(a-b)³ = a³ - 3a²b + 3ab² - b³]
\(\Rightarrow\) x³ - 4x² - x + 1 = x³ - 6x² + 12x - 8
\(\Rightarrow\) x³ - 4x² - x + 1 - x³ + 6x² - 12x + 8 = 0
\(\Rightarrow\) 2x² - 13x + 9 = 0
which is of the form ax² + bx + c = 0, a \(\neq\)0
\(\therefore\) It is a quadratic equation.

Given, equation is (x+2)³=2x(x²-1)
\(\Rightarrow\) x³ + 8 + 3x²(2) + 3x (2)² = 2x³ - 2x
[\(\because\) (a + b)³ = a³ + b³ + 3a²b + 3ab²]
\(\Rightarrow\) x³ + 8 + 6x² + 12x = 2x³ - 2x
\(\Rightarrow\) x³ + 8 + 6x² + 12x - 2x³ + 2x = 0
\(\Rightarrow\) -x³ + 6x² + 14x + 8 = 0
which is not of the form ax² + bx + c = 0,
because it has cubic term, i.e. x³.
\(\therefore\) It is not a quadratic equation.

(2x-1)(x-3)=(x+5)(x-1)
2x²-6x-x+3=x²-x+5x-5
2x²-6x-x+3-x²+x-5x+5=0
x²-11x+8=0
Which is of the form ax²+bx+c=0. Hence the given equation is a quadratic equation.

Given equation (x-3)(2x+1)=x(x+5)
2x² + x - 6x - 3 = x² + 5x
\(\Rightarrow\) 2x² + x - 6x - 3 - x² - 5x = 0
\(\Rightarrow\) x² - 10x - 3 = 0 has its highest power 2
Hence it is a quadratic equation.

We have
6x² – x – 2 = 6x² + 3x – 4x – 2
= 3x (2x + 1) – 2 (2x + 1)
= (3x – 2)(2x + 1)
The roots of 6x² - x - 2 = 0 are the values of x for which (3x - 2)(2x + 1) = 0
Therefore, 3x - 2 = 0 or 2x + 1 = 0,
ie., x = \(\frac{2}{3}\) or x = \(-\frac{1}{2}\)
Therefore, the roots of 6x² - x - 2 = 0 are \(\frac{2}{3}\) and \(-\frac{1}{2}\)
We verify the roots, by checking that \(\frac{2}{3}\) and \(-\frac{1}{2}\) satisfy 6x² - x - 2 = 0.

\(3 x^{2}-2 \sqrt{6} x+2=3 x^{2}-\sqrt{6} x-\sqrt{6} x+2\)
\(=\sqrt{3} x(\sqrt{3} x-\sqrt{2})-\sqrt{2}(\sqrt{3} x-\sqrt{2})\)
\(=(\sqrt{3} x-\sqrt{2})(\sqrt{3} x-\sqrt{2})\)
So, the roots of the equation are the values of x for which
\((\sqrt{3} x-\sqrt{2})(\sqrt{3} x-\sqrt{2})=0\)
Now, \(\sqrt{3} x-\sqrt{2}=0 \text { for } x=\sqrt{\frac{2}{3}}\)
So, this root is repeated twice, one for each repeated factor \(\sqrt{3} x-\sqrt{2}\)
Therefore, the roots of 3x² - 2\(\sqrt6\)x + 2 = 0 are \(\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}}\)

Let the two consecutive integers be x and x+1
ATQ x²+(x+1)2=365
\(\Rightarrow\) x²+x2+2x+1=365 \(\Rightarrow\) 2x²+2x-364=0
\(\Rightarrow\) x²+x-182=0 \(\Rightarrow\) x²+14x-13x-182=0
\(\Rightarrow\) x(x+14)-13(x+14)=0 \(\Rightarrow\) (x-13)(x+14)=0
\(\Rightarrow\) x=13, -14 (-14 is rejected because it is a negative integer)
Hence, the two consecutive positive integers are 13 and 13+1=14