2x^2+16=-18x

Solution for 2x^2+16=-18x equation:

2x^2+16=-18x

We move all terms to the left:

2x^2+16-(-18x)=0

We get rid of parentheses

2x^2+18x+16=0

a = 2; b = 18; c = +16;
Δ = b2-4ac
Δ = 182-4·2·16
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-14}{2*2}=\frac{-32}{4}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+14}{2*2}=\frac{-4}{4}
=-1
$