What are the solutions to the equation (x-5)(x+1)=7 ?

The solutions to the equation (x-5)(x+1)=7 are x=6,x=-2

Find the zeros of (x-5)(x+1)=7

The zeros of (x-5)(x+1)=7 are x=6,x=-2

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(x - 5) (x + 1) = 7

x = 6, x = - 2

Solution steps

(x - 5) (x + 1) = 7

Expand

(x - 5) (x + 1) : x^{2} - 4 x - 5

x^{2} - 4 x - 5 = 7

Move

7

to the left side

x^{2} - 4 x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 12) = 8

x_{1, 2} = \frac{- ( - 4 ) \pm 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 8}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 8}{2 \cdot 1}

x = \frac{- ( - 4 ) + 8}{2 \cdot 1} : 6

x = \frac{- ( - 4 ) - 8}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 6, x = - 2

x^{2} - 9 x + 25 = 5

a = π r^{2}

x, x^{2} + y^{2} + 2 x - 4 y + 5 = 0

\frac{p ^{2} + 21}{7} = 10

0 = x^{2} + 4

What are the solutions to the equation (x-5)(x+1)=7 ?
The solutions to the equation (x-5)(x+1)=7 are x=6,x=-2
Find the zeros of (x-5)(x+1)=7
The zeros of (x-5)(x+1)=7 are x=6,x=-2