13=-16x^2+32x+5

Solution for 13=-16x^2+32x+5 equation:

13=-16x^2+32x+5

We move all terms to the left:

13-(-16x^2+32x+5)=0

We get rid of parentheses

16x^2-32x-5+13=0

We add all the numbers together, and all the variables

16x^2-32x+8=0

a = 16; b = -32; c = +8;
Δ = b2-4ac
Δ = -322-4·16·8
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16\sqrt{2}}{2*16}=\frac{32-16\sqrt{2}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16\sqrt{2}}{2*16}=\frac{32+16\sqrt{2}}{32}
$