225x^2-150x+1=0

Solution for 225x^2-150x+1=0 equation:

225x^2-150x+1=0

a = 225; b = -150; c = +1;
Δ = b2-4ac
Δ = -1502-4·225·1
Δ = 21600
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:

$\sqrt{\Delta}=\sqrt{21600}=\sqrt{3600*6}=\sqrt{3600}*\sqrt{6}=60\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-150)-60\sqrt{6}}{2*225}=\frac{150-60\sqrt{6}}{450}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-150)+60\sqrt{6}}{2*225}=\frac{150+60\sqrt{6}}{450}
$