-x^2+x+96=0

Solution for -x^2+x+96=0 equation:

-x^2+x+96=0

We add all the numbers together, and all the variables

-1x^2+x+96=0

a = -1; b = 1; c = +96;
Δ = b2-4ac
Δ = 12-4·(-1)·96
Δ = 385
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{-(1)-\sqrt{385}}{2*-1}=\frac{-1-\sqrt{385}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{385}}{2*-1}=\frac{-1+\sqrt{385}}{-2}
$