300=8x^2+16x

Solution for 300=8x^2+16x equation:

300=8x^2+16x

We move all terms to the left:

300-(8x^2+16x)=0

We get rid of parentheses

-8x^2-16x+300=0

a = -8; b = -16; c = +300;
Δ = b2-4ac
Δ = -162-4·(-8)·300
Δ = 9856
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:

$\sqrt{\Delta}=\sqrt{9856}=\sqrt{64*154}=\sqrt{64}*\sqrt{154}=8\sqrt{154}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{154}}{2*-8}=\frac{16-8\sqrt{154}}{-16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{154}}{2*-8}=\frac{16+8\sqrt{154}}{-16}
$