0=2x^2+16x-

Solution for 0=2x^2+16x- equation:

0=2x^2+16x-

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-2x^2-16x+=0

We add all the numbers together, and all the variables

-2x^2-16x=0

a = -2; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·(-2)·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*-2}=\frac{0}{-4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*-2}=\frac{32}{-4}
=-8
$