x=-16x^2+42x+120

Solution for x=-16x^2+42x+120 equation:

x=-16x^2+42x+120

We move all terms to the left:

x-(-16x^2+42x+120)=0

We get rid of parentheses

16x^2-42x+x-120=0

We add all the numbers together, and all the variables

16x^2-41x-120=0

a = 16; b = -41; c = -120;
Δ = b2-4ac
Δ = -412-4·16·(-120)
Δ = 9361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-\sqrt{9361}}{2*16}=\frac{41-\sqrt{9361}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+\sqrt{9361}}{2*16}=\frac{41+\sqrt{9361}}{32}
$