150=(-16x^2)+120x+3

Solution for 150=(-16x^2)+120x+3 equation:

150=(-16x^2)+120x+3

We move all terms to the left:

150-((-16x^2)+120x+3)=0


We calculate terms in parentheses: -((-16x^2)+120x+3), so:

(-16x^2)+120x+3

We get rid of parentheses

-16x^2+120x+3

Back to the equation:

-(-16x^2+120x+3)


We get rid of parentheses

16x^2-120x-3+150=0

We add all the numbers together, and all the variables

16x^2-120x+147=0

a = 16; b = -120; c = +147;
Δ = b2-4ac
Δ = -1202-4·16·147
Δ = 4992
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:

$\sqrt{\Delta}=\sqrt{4992}=\sqrt{64*78}=\sqrt{64}*\sqrt{78}=8\sqrt{78}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-8\sqrt{78}}{2*16}=\frac{120-8\sqrt{78}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+8\sqrt{78}}{2*16}=\frac{120+8\sqrt{78}}{32}
$