3x^2+42x-99=0

Solution for 3x^2+42x-99=0 equation:

3x^2+42x-99=0

a = 3; b = 42; c = -99;
Δ = b2-4ac
Δ = 422-4·3·(-99)
Δ = 2952
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:

$\sqrt{\Delta}=\sqrt{2952}=\sqrt{36*82}=\sqrt{36}*\sqrt{82}=6\sqrt{82}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-6\sqrt{82}}{2*3}=\frac{-42-6\sqrt{82}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+6\sqrt{82}}{2*3}=\frac{-42+6\sqrt{82}}{6}
$