17x^2+32x-16=0

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

17x^2+32x-16=0

a = 17; b = 32; c = -16;
Δ = b2-4ac
Δ = 322-4·17·(-16)
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*17}=\frac{-32-8\sqrt{33}}{34}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8\sqrt{33}}{2*17}=\frac{-32+8\sqrt{33}}{34}
$