12=16x^2+32x

Solution for 12=16x^2+32x equation:

12=16x^2+32x

We move all terms to the left:

12-(16x^2+32x)=0

We get rid of parentheses

-16x^2-32x+12=0

a = -16; b = -32; c = +12;
Δ = b2-4ac
Δ = -322-4·(-16)·12
Δ = 1792
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1792}=\sqrt{256*7}=\sqrt{256}*\sqrt{7}=16\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16\sqrt{7}}{2*-16}=\frac{32-16\sqrt{7}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16\sqrt{7}}{2*-16}=\frac{32+16\sqrt{7}}{-32}
$