9=2x^2+16x

Solution for 9=2x^2+16x equation:

9=2x^2+16x

We move all terms to the left:

9-(2x^2+16x)=0

We get rid of parentheses

-2x^2-16x+9=0

a = -2; b = -16; c = +9;
Δ = b2-4ac
Δ = -162-4·(-2)·9
Δ = 328
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:

$\sqrt{\Delta}=\sqrt{328}=\sqrt{4*82}=\sqrt{4}*\sqrt{82}=2\sqrt{82}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{82}}{2*-2}=\frac{16-2\sqrt{82}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{82}}{2*-2}=\frac{16+2\sqrt{82}}{-4}
$