0=4x^2+20x-26

Solution for 0=4x^2+20x-26 equation:

0=4x^2+20x-26

We move all terms to the left:

0-(4x^2+20x-26)=0

We add all the numbers together, and all the variables

-(4x^2+20x-26)=0

We get rid of parentheses

-4x^2-20x+26=0

a = -4; b = -20; c = +26;
Δ = b2-4ac
Δ = -202-4·(-4)·26
Δ = 816
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:

$\sqrt{\Delta}=\sqrt{816}=\sqrt{16*51}=\sqrt{16}*\sqrt{51}=4\sqrt{51}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{51}}{2*-4}=\frac{20-4\sqrt{51}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{51}}{2*-4}=\frac{20+4\sqrt{51}}{-8}
$