4x^2+20x=13

Solution for 4x^2+20x=13 equation:

4x^2+20x=13

We move all terms to the left:

4x^2+20x-(13)=0

a = 4; b = 20; c = -13;
Δ = b2-4ac
Δ = 202-4·4·(-13)
Δ = 608
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:

$\sqrt{\Delta}=\sqrt{608}=\sqrt{16*38}=\sqrt{16}*\sqrt{38}=4\sqrt{38}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{38}}{2*4}=\frac{-20-4\sqrt{38}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{38}}{2*4}=\frac{-20+4\sqrt{38}}{8}
$