**Solving Quadratic Equations by Quadratic Formula Examples :**

Here we are going to see how we use the formula to solve a quadratic equation.

The formula to find a quadratic equation ax^{2} + bx + c = 0 is x =

**Question 1 :**

Solve the following quadratic equations by formula method

(i) 2x^{2} −5x + 2 = 0

**Solution :**

By comparing the given quadratic equation with the general form of quadratic equation ax^{2} + bx + c = 0, we get

a = 2, b = -5 and c = 2

x = [-b ± √b^{2} - 4ac]/2a

x = [-(-5) ± √(-5)^{2} - 4(2)(2)]/2(2)

x = [5 ± √(25 - 16)]/4

x = [5 ± √9]/4

x = [5 ± 3]/4

x = (5 + 3) /4 and x = (5 - 3)/4

x = 8/4 and x = 2/4

x = 2 and x = 1/2.

Hence the solutions are 2 and 1/2.

Let us look into the next problem on "Solving Quadratic Equations by Quadratic Formula Examples".

(ii) √2f^{2} −6f + 3√2 = 0

**Solution :**

By comparing the given quadratic equation with the general form of quadratic equation ax^{2} + bx + c = 0, we get

a = √2, b = -6 and c = 3√2

x = [-b ± √b^{2} - 4ac]/2a

x = [-(-6) ± √(-6)^{2} - 4(√2)(3√2)]/2(√2)

x = [6 ± √(36 - 24)]/2√2

x = [6 ± √12]/2√2

x = [6 ± 2√3]/2√2

x = [6 + 2√3]/2√2 x = 2(3 + √3)/2√2 x = (3 + √3)/√2 |
x = [6 - 2√3]/2√2 x = 2(3 - √3)/2√2 x = (3 - √3)/√2 |

Hence the solutions are (3 + √3)/√2 and (3 - √3)/√2.

(iii) 3y^{2} −20y - 23 = 0

**Solution :**

By comparing the given quadratic equation with the general form of quadratic equation ax^{2} + bx + c = 0, we get

a = 3, b = -20 and c = -23

x = [-b ± √b^{2} - 4ac]/2a

x = [-(-20) ± √(-20)^{2} - 4(3)(-23)]/2(3)

x = [20 ± √(400 + 276)]/6

x = [20 ± √676]/6

x = [20 ± 26]/6

x = [20 + 26]/6 x = 46/6 x = 23/3 |
x = [20 - 26]/6 x = -6/6 x = -1 |

Hence the solutions are 23/3 and -1.

(iv) 36y^{2} −12ay + (a^{2} −b^{2}) = 0

**Solution :**

^{2} + bx + c = 0, we get

a = 36, b = -12a and c = (a^{2} −b^{2})

x = [-b ± √b^{2} - 4ac]/2a

x = [-(-12a) ± √(-12a)^{2} - 4(36)(a^{2} −b^{2})]/2(36)

x = [12a ± √(144a^{2} - 144a^{2} + 144b^{2})]/72

x = [12a ± √144b^{2}]/72

x = [12a ± 12b]/72

x = 12(a ± b)/72

x = (a ± b)/6

x = (a + b)/6 |
x = (a - b)/6 |

Hence the solutions are (a + b)/6 and (a - b)/6

**Question 2 :**

A ball rolls down a slope and travels a distance d = t^{2} −0.75t feet in t seconds. Find the time when the distance travelled by the ball is 11.25 feet.

**Solution :**

Distance travelled by the ball = 11.25 feet.

d = 11.25

11.25 = t^{2} −0.75t

t^{2} −0.75t - 11.25 = 0

t^{2} −0.75t = 11.25

Multiply the entire equation by 100

100t^{2} −75t = 1125

100t^{2} −75t - 1125 = 0

Divide by 25, we get

4t^{2} - 3t - 45 = 0

4t^{2} + 12t - 15t - 45 = 0

4t(t - 3) - 15(t + 3) = 0

(4t - 15) (t + 3) = 0

4t = 15 and t = -3

t = 15/4 = 3.75 seconds

After having gone through the stuff given above, we hope that the students would have understood, "Solving Quadratic Equations by Quadratic Formula Examples".

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