0=-16x^2+32x+17

Solution for 0=-16x^2+32x+17 equation:

0=-16x^2+32x+17

We move all terms to the left:

0-(-16x^2+32x+17)=0

We add all the numbers together, and all the variables

-(-16x^2+32x+17)=0

We get rid of parentheses

16x^2-32x-17=0

a = 16; b = -32; c = -17;
Δ = b2-4ac
Δ = -322-4·16·(-17)
Δ = 2112
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:

$\sqrt{\Delta}=\sqrt{2112}=\sqrt{64*33}=\sqrt{64}*\sqrt{33}=8\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8\sqrt{33}}{2*16}=\frac{32-8\sqrt{33}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8\sqrt{33}}{2*16}=\frac{32+8\sqrt{33}}{32}
$