2x^2+16x=-18

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

2x^2+16x=-18

We move all terms to the left:

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

We add all the numbers together, and all the variables

2x^2+16x+18=0

a = 2; b = 16; c = +18;
Δ = b2-4ac
Δ = 162-4·2·18
Δ = 112
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{7}}{2*2}=\frac{-16-4\sqrt{7}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{7}}{2*2}=\frac{-16+4\sqrt{7}}{4}
$