x2+16x-96=0

Solution for x2+16x-96=0 equation:

x2+16x-96=0

We add all the numbers together, and all the variables

x^2+16x-96=0

a = 1; b = 16; c = -96;
Δ = b2-4ac
Δ = 162-4·1·(-96)
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{10}}{2*1}=\frac{-16-8\sqrt{10}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{10}}{2*1}=\frac{-16+8\sqrt{10}}{2}
$