0=4x^2+26x-160

Solution for 0=4x^2+26x-160 equation:

0=4x^2+26x-160

We move all terms to the left:

0-(4x^2+26x-160)=0

We add all the numbers together, and all the variables

-(4x^2+26x-160)=0

We get rid of parentheses

-4x^2-26x+160=0

a = -4; b = -26; c = +160;
Δ = b2-4ac
Δ = -262-4·(-4)·160
Δ = 3236
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:

$\sqrt{\Delta}=\sqrt{3236}=\sqrt{4*809}=\sqrt{4}*\sqrt{809}=2\sqrt{809}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-2\sqrt{809}}{2*-4}=\frac{26-2\sqrt{809}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2\sqrt{809}}{2*-4}=\frac{26+2\sqrt{809}}{-8}
$