5x^2-40x+13=0

Solution for 5x^2-40x+13=0 equation:

5x^2-40x+13=0

a = 5; b = -40; c = +13;
Δ = b2-4ac
Δ = -402-4·5·13
Δ = 1340
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:

$\sqrt{\Delta}=\sqrt{1340}=\sqrt{4*335}=\sqrt{4}*\sqrt{335}=2\sqrt{335}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-2\sqrt{335}}{2*5}=\frac{40-2\sqrt{335}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+2\sqrt{335}}{2*5}=\frac{40+2\sqrt{335}}{10}
$