x^2+20x-18=3

Solution for x^2+20x-18=3 equation:

x^2+20x-18=3

We move all terms to the left:

x^2+20x-18-(3)=0

We add all the numbers together, and all the variables

x^2+20x-21=0

a = 1; b = 20; c = -21;
Δ = b2-4ac
Δ = 202-4·1·(-21)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-22}{2*1}=\frac{-42}{2}
=-21
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+22}{2*1}=\frac{2}{2}
=1
$