2r^2+12r=64

Solution for 2r^2+12r=64 equation:

2r^2+12r=64

We move all terms to the left:

2r^2+12r-(64)=0

a = 2; b = 12; c = -64;
Δ = b2-4ac
Δ = 122-4·2·(-64)
Δ = 656
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:

$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{41}}{2*2}=\frac{-12-4\sqrt{41}}{4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{41}}{2*2}=\frac{-12+4\sqrt{41}}{4}
$