3x^2+33x-48=0

Solution for 3x^2+33x-48=0 equation:

3x^2+33x-48=0

a = 3; b = 33; c = -48;
Δ = b2-4ac
Δ = 332-4·3·(-48)
Δ = 1665
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:

$\sqrt{\Delta}=\sqrt{1665}=\sqrt{9*185}=\sqrt{9}*\sqrt{185}=3\sqrt{185}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-3\sqrt{185}}{2*3}=\frac{-33-3\sqrt{185}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+3\sqrt{185}}{2*3}=\frac{-33+3\sqrt{185}}{6}
$