What are the solutions to the equation (x-2)(x+9)=42 ?

The solutions to the equation (x-2)(x+9)=42 are x=5,x=-12

Find the zeros of (x-2)(x+9)=42

The zeros of (x-2)(x+9)=42 are x=5,x=-12

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(x - 2) (x + 9) = 42

x = 5, x = - 12

Solution steps

(x - 2) (x + 9) = 42

Expand

(x - 2) (x + 9) : x^{2} + 7 x - 18

x^{2} + 7 x - 18 = 42

Move

42

to the left side

x^{2} + 7 x - 60 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot ( - 60 )}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot (- 60) = 17

x_{1, 2} = \frac{- 7 \pm 17}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 7 + 17}{2 \cdot 1}, x_{2} = \frac{- 7 - 17}{2 \cdot 1}

x = \frac{- 7 + 17}{2 \cdot 1} : 5

x = \frac{- 7 - 17}{2 \cdot 1} : - 12

The solutions to the quadratic equation are:

x = 5, x = - 12

2 k^{2} + 11 k + 14 = 0

y, x^{3} + y^{2} = 4 + x y

24 x^{2} - 48 x = 0

s^{2} - s - 1

x^{2} = 5 + x

What are the solutions to the equation (x-2)(x+9)=42 ?
The solutions to the equation (x-2)(x+9)=42 are x=5,x=-12
Find the zeros of (x-2)(x+9)=42
The zeros of (x-2)(x+9)=42 are x=5,x=-12