4=-16x^2+32x+16

Solution for 4=-16x^2+32x+16 equation:

4=-16x^2+32x+16

We move all terms to the left:

4-(-16x^2+32x+16)=0

We get rid of parentheses

16x^2-32x-16+4=0

We add all the numbers together, and all the variables

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}
$