4x^2+20=-18x

Solution for 4x^2+20=-18x equation:

4x^2+20=-18x

We move all terms to the left:

4x^2+20-(-18x)=0

We get rid of parentheses

4x^2+18x+20=0

a = 4; b = 18; c = +20;
Δ = b2-4ac
Δ = 182-4·4·20
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2}{2*4}=\frac{-20}{8}
=-2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2}{2*4}=\frac{-16}{8}
=-2
$