13x^2+32x-360=0

Solution for 13x^2+32x-360=0 equation:

13x^2+32x-360=0

a = 13; b = 32; c = -360;
Δ = b2-4ac
Δ = 322-4·13·(-360)
Δ = 19744
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:

$\sqrt{\Delta}=\sqrt{19744}=\sqrt{16*1234}=\sqrt{16}*\sqrt{1234}=4\sqrt{1234}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4\sqrt{1234}}{2*13}=\frac{-32-4\sqrt{1234}}{26}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4\sqrt{1234}}{2*13}=\frac{-32+4\sqrt{1234}}{26}
$