5x^2-40x+17=0

Solution for 5x^2-40x+17=0 equation:

5x^2-40x+17=0

a = 5; b = -40; c = +17;
Δ = b2-4ac
Δ = -402-4·5·17
Δ = 1260
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:

$\sqrt{\Delta}=\sqrt{1260}=\sqrt{36*35}=\sqrt{36}*\sqrt{35}=6\sqrt{35}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-6\sqrt{35}}{2*5}=\frac{40-6\sqrt{35}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+6\sqrt{35}}{2*5}=\frac{40+6\sqrt{35}}{10}
$