3x+2x^2-99=0

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

3x+2x^2-99=0

a = 2; b = 3; c = -99;
Δ = b2-4ac
Δ = 32-4·2·(-99)
Δ = 801
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:

$\sqrt{\Delta}=\sqrt{801}=\sqrt{9*89}=\sqrt{9}*\sqrt{89}=3\sqrt{89}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{89}}{2*2}=\frac{-3-3\sqrt{89}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{89}}{2*2}=\frac{-3+3\sqrt{89}}{4}
$