2x2-20x+9=0

Solution for 2x2-20x+9=0 equation:

2x^2-20x+9=0

a = 2; b = -20; c = +9;
Δ = b2-4ac
Δ = -202-4·2·9
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*2}=\frac{20-2\sqrt{82}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{82}}{2*2}=\frac{20+2\sqrt{82}}{4}
$