8r^2+16r=40

Solution for 8r^2+16r=40 equation:

8r^2+16r=40

We move all terms to the left:

8r^2+16r-(40)=0

a = 8; b = 16; c = -40;
Δ = b2-4ac
Δ = 162-4·8·(-40)
Δ = 1536
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:

$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{6}}{2*8}=\frac{-16-16\sqrt{6}}{16}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{6}}{2*8}=\frac{-16+16\sqrt{6}}{16}
$