5x^2+20x+13=0

Solution for 5x^2+20x+13=0 equation:

5x^2+20x+13=0

a = 5; b = 20; c = +13;
Δ = b2-4ac
Δ = 202-4·5·13
Δ = 140
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:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{35}}{2*5}=\frac{-20-2\sqrt{35}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{35}}{2*5}=\frac{-20+2\sqrt{35}}{10}
$