4x^2+52x-160=0

Solution for 4x^2+52x-160=0 equation:

4x^2+52x-160=0

a = 4; b = 52; c = -160;
Δ = b2-4ac
Δ = 522-4·4·(-160)
Δ = 5264
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:

$\sqrt{\Delta}=\sqrt{5264}=\sqrt{16*329}=\sqrt{16}*\sqrt{329}=4\sqrt{329}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(52)-4\sqrt{329}}{2*4}=\frac{-52-4\sqrt{329}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(52)+4\sqrt{329}}{2*4}=\frac{-52+4\sqrt{329}}{8}
$