3x2+9x-6=0

Solution for 3x2+9x-6=0 equation:

3x^2+9x-6=0

a = 3; b = 9; c = -6;
Δ = b2-4ac
Δ = 92-4·3·(-6)
Δ = 153
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:

$\sqrt{\Delta}=\sqrt{153}=\sqrt{9*17}=\sqrt{9}*\sqrt{17}=3\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{17}}{2*3}=\frac{-9-3\sqrt{17}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{17}}{2*3}=\frac{-9+3\sqrt{17}}{6}
$