17x^2+92x-160=0

Solution for 17x^2+92x-160=0 equation:

17x^2+92x-160=0

a = 17; b = 92; c = -160;
Δ = b2-4ac
Δ = 922-4·17·(-160)
Δ = 19344
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:

$\sqrt{\Delta}=\sqrt{19344}=\sqrt{16*1209}=\sqrt{16}*\sqrt{1209}=4\sqrt{1209}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(92)-4\sqrt{1209}}{2*17}=\frac{-92-4\sqrt{1209}}{34}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(92)+4\sqrt{1209}}{2*17}=\frac{-92+4\sqrt{1209}}{34}
$