X^2-20x+18=0

Solution for X^2-20x+18=0 equation:

X^2-20X+18=0

a = 1; b = -20; c = +18;
Δ = b2-4ac
Δ = -202-4·1·18
Δ = 328
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:

$\sqrt{\Delta}=\sqrt{328}=\sqrt{4*82}=\sqrt{4}*\sqrt{82}=2\sqrt{82}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{82}}{2*1}=\frac{20-2\sqrt{82}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{82}}{2*1}=\frac{20+2\sqrt{82}}{2}
$