x^2-20x+93=0

Solution for x^2-20x+93=0 equation:

x^2-20x+93=0

a = 1; b = -20; c = +93;
Δ = b2-4ac
Δ = -202-4·1·93
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{7}}{2*1}=\frac{20-2\sqrt{7}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{7}}{2*1}=\frac{20+2\sqrt{7}}{2}
$