x^2+4x-5=99

Solution for x^2+4x-5=99 equation:

x^2+4x-5=99

We move all terms to the left:

x^2+4x-5-(99)=0

We add all the numbers together, and all the variables

x^2+4x-104=0

a = 1; b = 4; c = -104;
Δ = b2-4ac
Δ = 42-4·1·(-104)
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12\sqrt{3}}{2*1}=\frac{-4-12\sqrt{3}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12\sqrt{3}}{2*1}=\frac{-4+12\sqrt{3}}{2}
$