j^2-18j=23

Solution for j^2-18j=23 equation:

j^2-18j=23

We move all terms to the left:

j^2-18j-(23)=0

a = 1; b = -18; c = -23;
Δ = b2-4ac
Δ = -182-4·1·(-23)
Δ = 416
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{416}=\sqrt{16*26}=\sqrt{16}*\sqrt{26}=4\sqrt{26}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-4\sqrt{26}}{2*1}=\frac{18-4\sqrt{26}}{2}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+4\sqrt{26}}{2*1}=\frac{18+4\sqrt{26}}{2}
$