x^2+23x=41

Solution for x^2+23x=41 equation:

x^2+23x=41

We move all terms to the left:

x^2+23x-(41)=0

a = 1; b = 23; c = -41;
Δ = b2-4ac
Δ = 232-4·1·(-41)
Δ = 693
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:

$\sqrt{\Delta}=\sqrt{693}=\sqrt{9*77}=\sqrt{9}*\sqrt{77}=3\sqrt{77}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-3\sqrt{77}}{2*1}=\frac{-23-3\sqrt{77}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+3\sqrt{77}}{2*1}=\frac{-23+3\sqrt{77}}{2}
$