Y=-16x^2+120+3

Solution for Y=-16x^2+120+3 equation:

=-16Y^2+120+3

We move all terms to the left:

-(-16Y^2+120+3)=0

We get rid of parentheses

16Y^2-120-3=0

We add all the numbers together, and all the variables

16Y^2-123=0

a = 16; b = 0; c = -123;
Δ = b2-4ac
Δ = 02-4·16·(-123)
Δ = 7872
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{7872}=\sqrt{64*123}=\sqrt{64}*\sqrt{123}=8\sqrt{123}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{123}}{2*16}=\frac{0-8\sqrt{123}}{32}
=-\frac{8\sqrt{123}}{32}
=-\frac{\sqrt{123}}{4}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{123}}{2*16}=\frac{0+8\sqrt{123}}{32}
=\frac{8\sqrt{123}}{32}
=\frac{\sqrt{123}}{4}
$