a^2-14a-96=0

Solution for a^2-14a-96=0 equation:

a^2-14a-96=0

a = 1; b = -14; c = -96;
Δ = b2-4ac
Δ = -142-4·1·(-96)
Δ = 580
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{580}=\sqrt{4*145}=\sqrt{4}*\sqrt{145}=2\sqrt{145}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{145}}{2*1}=\frac{14-2\sqrt{145}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{145}}{2*1}=\frac{14+2\sqrt{145}}{2}
$