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

Solution for (x-5)(x+1)=7 equation:

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

We move all terms to the left:

(x-5)(x+1)-(7)=0

We multiply parentheses ..

(+x^2+x-5x-5)-7=0

We get rid of parentheses

x^2+x-5x-5-7=0

We add all the numbers together, and all the variables

x^2-4x-12=0

a = 1; b = -4; c = -12;
Δ = b2-4ac
Δ = -42-4·1·(-12)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*1}=\frac{12}{2}
=6
$