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.

Let the breadth of the park be x m.
Given, perimeter of a rectangular park = 80 m
\(\Rightarrow\) 2(Length + Breadth) = 80 m
\(\Rightarrow\) Length + Breadth = 40 m
\(\Rightarrow\) Length = (40 - x)m
Now, area of a rectangular park = Length \(\times\)Breadth
=(40 - x)x m²
But according to the question, area of the rectangular park is 400 m².
\(\therefore\) (40 - x) x = 400
\(\Rightarrow\) x² - 40x + 400 = 0
\(\Rightarrow\) x² - 2x \(\times\) 20 + (20)² = 0
\(\Rightarrow\) (x - 20)² = 0 [\(\because\) a² - 2ab + b² = (a - b)²]
\(\Rightarrow\) x = 20
thus, breadth of the park = 20 m
and length of the park = 40 - 20 = 20 m
Hence, it is possible to design the rectangular park having perimeter 80 m and area 400 m with equal length and breadth, i.e. 20 m each.

Let the age of one of two friends be x yr.
Then, age of other friend = (20 - x) yr.
[\(\because\) the sum of the ages of two friends is 20 yr]
4 yr ago, age of one of two friends = (x - 4) yr
and age of the other friend = (20 - x - 4) yr = (16 - x) yr
According to the question, (x - 4)(16-x) = 48
\(\Rightarrow\) 16x - x² - 64 + 4x = 48
\(\Rightarrow\) x² - 20x + 112 = 0
On comparing with ax² + bx + c = 0, we get
a = 1, b = -20 and c = 112
Now, discriminant, D = b² - 4ac
=(-20)² - 4 \(\times\) 1 \(\times\) 112
=400 - 448 = -48 < 0
which implies that the real roots are not possible because this condition represents imaginary roots. So, the solution does not exist and hence given situation is not possible.

Let breadth of a rectangular mango grove be x m
Then, length of a rectangular mango grove = 2x m
According to the question,
Area of rectangular mango grove = 800 m²
\(\Rightarrow\) 2x(x) = 800  [\(\because\) Area = length \(\times\) breadth]
\(\Rightarrow\) 2x² = 800 \(\Rightarrow\) x² = 400
\(\Rightarrow\) x = \(\pm\)20
But, x = -20 is not possible because breadth can never be negative. so, x = 20.
Thus, length = 2x = 40 m and breadth = 20 m.

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 \(\frac{2}{\sqrt{3}} \text { and } \frac{2}{\sqrt{3}}\)

Consider the equation
x² - 3x + 5 = 0
Comparing it with ax² + bx + c = 0, we get
a = 2, b = -3 and c = 5
Discriminant = b2 - 4ac
= (-3)2 - 4 (2) (5) = 9 - 40
= - 31
As b² - 4ac < 0,
Therefore, no real root is possible for the given equation.

Let one number be x,
Then, another number be 27 - x
[\(\because\) sum of two numbers = 27]
According to the question,
Product of these two numbers = 182
\(\therefore\) x(27-x)=182
\(\Rightarrow\) 27x-x² = 182
\(\Rightarrow\) x²-27x+182=0 
\(\Rightarrow\) x²-14x-13x+182=0
[\(\because\) (-14) \(\times\) (-13) =182 and -14-13=-27
\(\Rightarrow\) x(x-14)-13(x-14)=0 \(\Rightarrow\) (x-14)(x-13)=0
\(\Rightarrow\) x-14 = 0 or x-13 = 0
\(\Rightarrow\) x=14 or x=13
If x = 14, then 27-x= 27-14 =13 and if x=13, then 27-x= 27-13 = 14 Hence, in both cases, the numbers are 13 and 14.

\(2x^2-x+{1\over 8}=0\)
\({16x^2-8x+1\over 8}=0\)
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 \(\frac{3}{2}\)

Let the two consecutive positive integers be x and x+1
Then, according to the question,
x(x+1) = 306 \(\Rightarrow\) x² + x = 306
\(\Rightarrow\) x² + x - 306 = 0,
Which is the required quadratic equation.

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.

9x²-9a(a+b)x+(2a²+5ab+2b²)=0
Here A=9, B=-9(a+b),C =2a²+5ab+2b²
D = B²-4AC =[-9(a+b]²-4X9(2a²+5ab+2b²) = 81 (a+b)² -36(2a²+5ab+2b²)
= 81(a²+b²+2ab)-72a²-180ab-72b²-180ab-72b²=9a²+9b²-18ab
= 9(a²+b²-2ab)=9(a-b)²
\(x=\frac { B\pm \sqrt { D } }{ 2A } =\frac { 9(a+b)\pm 3(a-b) }{ 18 } =\frac { 2a+b }{ 3 } ,\frac { a+2b }{ 3 } \)

Let the original speed of train be x km/hr. Then,
Increased speed of the train = (x + 5)km/hr
Time taken by the train under usual speed to cover 360 km = \(\frac{360}{x} \mathrm{hr}\)
Time taken by the train under increased speed to cover 360 km = \(\frac{360}{x+5} \mathrm{hr}\)
Therefore,
\( \frac{360}{x}-\frac{360}{x+5}=1 \)
\(\frac{360(x+5)-360 x}{x(x+5)}=1 \)
\(\frac{360 x+1800-360 x}{x^{2}+5 x}=1 \)
\(\frac{1800}{x^{2}+5 x}=1\)
1800 = x² + 5x
x² + 5x - 1800 = 0
x² - 40x + 45x - 1800 = 0
x(x - 40) + 45(x - 40) = 0
(x - 40)(x + 45) = 0
x - 40 = 0
x = 40
Or
x + 45 = 0
x = -45
But, the speed of the train can never be negative.
Hence, the original speed of train is x = 40 km/hr.