(r-4)(r+8)=-24

Solution for (r-4)(r+8)=-24 equation:

(r-4)(r+8)=-24

We move all terms to the left:

(r-4)(r+8)-(-24)=0

We add all the numbers together, and all the variables

(r-4)(r+8)+24=0

We multiply parentheses ..

(+r^2+8r-4r-32)+24=0

We get rid of parentheses

r^2+8r-4r-32+24=0

We add all the numbers together, and all the variables

r^2+4r-8=0

a = 1; b = 4; c = -8;
Δ = b2-4ac
Δ = 42-4·1·(-8)
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{3}}{2*1}=\frac{-4-4\sqrt{3}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{3}}{2*1}=\frac{-4+4\sqrt{3}}{2}
$