y^2+26y-160=0

Solution for y^2+26y-160=0 equation:

y^2+26y-160=0

a = 1; b = 26; c = -160;
Δ = b2-4ac
Δ = 262-4·1·(-160)
Δ = 1316
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1316}=\sqrt{4*329}=\sqrt{4}*\sqrt{329}=2\sqrt{329}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{329}}{2*1}=\frac{-26-2\sqrt{329}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{329}}{2*1}=\frac{-26+2\sqrt{329}}{2}
$