23x^2-124x+63=0

Solution for 23x^2-124x+63=0 equation:

23x^2-124x+63=0

a = 23; b = -124; c = +63;
Δ = b2-4ac
Δ = -1242-4·23·63
Δ = 9580
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:

$\sqrt{\Delta}=\sqrt{9580}=\sqrt{4*2395}=\sqrt{4}*\sqrt{2395}=2\sqrt{2395}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-124)-2\sqrt{2395}}{2*23}=\frac{124-2\sqrt{2395}}{46}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-124)+2\sqrt{2395}}{2*23}=\frac{124+2\sqrt{2395}}{46}
$