6x2+18x-3=0

Solution for 6x2+18x-3=0 equation:

6x^2+18x-3=0

a = 6; b = 18; c = -3;
Δ = b2-4ac
Δ = 182-4·6·(-3)
Δ = 396
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:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{11}}{2*6}=\frac{-18-6\sqrt{11}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{11}}{2*6}=\frac{-18+6\sqrt{11}}{12}
$