What are the solutions to the equation (x-2)(x+4)=7 ?

The solutions to the equation (x-2)(x+4)=7 are x=3,x=-5

Find the zeros of (x-2)(x+4)=7

The zeros of (x-2)(x+4)=7 are x=3,x=-5

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(x - 2) (x + 4) = 7

x = 3, x = - 5

Solution steps

(x - 2) (x + 4) = 7

Expand

(x - 2) (x + 4) : x^{2} + 2 x - 8

x^{2} + 2 x - 8 = 7

Move

7

to the left side

x^{2} + 2 x - 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 15 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 15) = 8

x_{1, 2} = \frac{- 2 \pm 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 8}{2 \cdot 1}, x_{2} = \frac{- 2 - 8}{2 \cdot 1}

x = \frac{- 2 + 8}{2 \cdot 1} : 3

x = \frac{- 2 - 8}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = 3, x = - 5

(x + 2)^{2} - 64 = 0

2 p = \frac{- 5 p - 6}{3}

3^{3}

6.5 (x - 6) = 19.5

3 x^{3} + 375

What are the solutions to the equation (x-2)(x+4)=7 ?
The solutions to the equation (x-2)(x+4)=7 are x=3,x=-5
Find the zeros of (x-2)(x+4)=7
The zeros of (x-2)(x+4)=7 are x=3,x=-5