300=-16x^2+128x+300

Solution for 300=-16x^2+128x+300 equation:

300=-16x^2+128x+300

We move all terms to the left:

300-(-16x^2+128x+300)=0

We get rid of parentheses

16x^2-128x-300+300=0

We add all the numbers together, and all the variables

16x^2-128x=0

a = 16; b = -128; c = 0;
Δ = b2-4ac
Δ = -1282-4·16·0
Δ = 16384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16384}=128$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-128}{2*16}=\frac{0}{32}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+128}{2*16}=\frac{256}{32}
=8
$