100X^2-500x+9=0

Solution for 100X^2-500x+9=0 equation:

100X^2-500X+9=0

a = 100; b = -500; c = +9;
Δ = b2-4ac
Δ = -5002-4·100·9
Δ = 246400
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:

$\sqrt{\Delta}=\sqrt{246400}=\sqrt{1600*154}=\sqrt{1600}*\sqrt{154}=40\sqrt{154}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-500)-40\sqrt{154}}{2*100}=\frac{500-40\sqrt{154}}{200}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-500)+40\sqrt{154}}{2*100}=\frac{500+40\sqrt{154}}{200}
$