4x^2+14x-99=0

Solution for 4x^2+14x-99=0 equation:

4x^2+14x-99=0

a = 4; b = 14; c = -99;
Δ = b2-4ac
Δ = 142-4·4·(-99)
Δ = 1780
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:

$\sqrt{\Delta}=\sqrt{1780}=\sqrt{4*445}=\sqrt{4}*\sqrt{445}=2\sqrt{445}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{445}}{2*4}=\frac{-14-2\sqrt{445}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{445}}{2*4}=\frac{-14+2\sqrt{445}}{8}
$