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

We add all the numbers together, and all the variables

j^2-18j+23=0

a = 1; b = -18; c = +23;
Δ = b2-4ac
Δ = -182-4·1·23
Δ = 232
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:

$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{58}}{2*1}=\frac{18-2\sqrt{58}}{2}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{58}}{2*1}=\frac{18+2\sqrt{58}}{2}
$