What are the solutions to the equation (x-5)(x+6)=12 ?

The solutions to the equation (x-5)(x+6)=12 are x=6,x=-7

Find the zeros of (x-5)(x+6)=12

The zeros of (x-5)(x+6)=12 are x=6,x=-7

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(x - 5) (x + 6) = 12

x = 6, x = - 7

Solution steps

(x - 5) (x + 6) = 12

Expand

(x - 5) (x + 6) : x^{2} + x - 30

x^{2} + x - 30 = 12

Move

12

to the left side

x^{2} + x - 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 42 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 42) = 13

x_{1, 2} = \frac{- 1 \pm 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 13}{2 \cdot 1}, x_{2} = \frac{- 1 - 13}{2 \cdot 1}

x = \frac{- 1 + 13}{2 \cdot 1} : 6

x = \frac{- 1 - 13}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

x = 6, x = - 7

2 w^{2} - 11 w + 8 = 0

16 + 40 x + 25 x^{2}

4 x^{2} + 17 x + 14 = 0

- 3 + 3 x^{5}

5 x = 3 x - 10 + 2 x

What are the solutions to the equation (x-5)(x+6)=12 ?
The solutions to the equation (x-5)(x+6)=12 are x=6,x=-7
Find the zeros of (x-5)(x+6)=12
The zeros of (x-5)(x+6)=12 are x=6,x=-7