4x^2+60x+99=0

Solution for 4x^2+60x+99=0 equation:

4x^2+60x+99=0

a = 4; b = 60; c = +99;
Δ = b2-4ac
Δ = 602-4·4·99
Δ = 2016
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:

$\sqrt{\Delta}=\sqrt{2016}=\sqrt{144*14}=\sqrt{144}*\sqrt{14}=12\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-12\sqrt{14}}{2*4}=\frac{-60-12\sqrt{14}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+12\sqrt{14}}{2*4}=\frac{-60+12\sqrt{14}}{8}
$