5x^2-62x+160=0

Solution for 5x^2-62x+160=0 equation:

5x^2-62x+160=0

a = 5; b = -62; c = +160;
Δ = b2-4ac
Δ = -622-4·5·160
Δ = 644
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:

$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-62)-2\sqrt{161}}{2*5}=\frac{62-2\sqrt{161}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-62)+2\sqrt{161}}{2*5}=\frac{62+2\sqrt{161}}{10}
$