-3r(r+8)=-40

Solution for -3r(r+8)=-40 equation:

-3r(r+8)=-40

We move all terms to the left:

-3r(r+8)-(-40)=0

We add all the numbers together, and all the variables

-3r(r+8)+40=0

We multiply parentheses

-3r^2-24r+40=0

a = -3; b = -24; c = +40;
Δ = b2-4ac
Δ = -242-4·(-3)·40
Δ = 1056
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:

$\sqrt{\Delta}=\sqrt{1056}=\sqrt{16*66}=\sqrt{16}*\sqrt{66}=4\sqrt{66}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-4\sqrt{66}}{2*-3}=\frac{24-4\sqrt{66}}{-6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+4\sqrt{66}}{2*-3}=\frac{24+4\sqrt{66}}{-6}
$