64=6y^2-32y

Solution for 64=6y^2-32y equation:

64=6y^2-32y

We move all terms to the left:

64-(6y^2-32y)=0

We get rid of parentheses

-6y^2+32y+64=0

a = -6; b = 32; c = +64;
Δ = b2-4ac
Δ = 322-4·(-6)·64
Δ = 2560
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16\sqrt{10}}{2*-6}=\frac{-32-16\sqrt{10}}{-12}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16\sqrt{10}}{2*-6}=\frac{-32+16\sqrt{10}}{-12}
$