13x^2+8x-96=0

Solution for 13x^2+8x-96=0 equation:

13x^2+8x-96=0

a = 13; b = 8; c = -96;
Δ = b2-4ac
Δ = 82-4·13·(-96)
Δ = 5056
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:

$\sqrt{\Delta}=\sqrt{5056}=\sqrt{64*79}=\sqrt{64}*\sqrt{79}=8\sqrt{79}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{79}}{2*13}=\frac{-8-8\sqrt{79}}{26}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{79}}{2*13}=\frac{-8+8\sqrt{79}}{26}
$