x^2=19x+18

Solution for x^2=19x+18 equation:

x^2=19x+18

We move all terms to the left:

x^2-(19x+18)=0

We get rid of parentheses

x^2-19x-18=0

a = 1; b = -19; c = -18;
Δ = b2-4ac
Δ = -192-4·1·(-18)
Δ = 433
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-\sqrt{433}}{2*1}=\frac{19-\sqrt{433}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{433}}{2*1}=\frac{19+\sqrt{433}}{2}
$