3x^2-16x-99=0

Solution for 3x^2-16x-99=0 equation:

3x^2-16x-99=0

a = 3; b = -16; c = -99;
Δ = b2-4ac
Δ = -162-4·3·(-99)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-38}{2*3}=\frac{-22}{6}
=-3+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+38}{2*3}=\frac{54}{6}
=9
$