3x^2+18x-69=0

Solution for 3x^2+18x-69=0 equation:

3x^2+18x-69=0

a = 3; b = 18; c = -69;
Δ = b2-4ac
Δ = 182-4·3·(-69)
Δ = 1152
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:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-24\sqrt{2}}{2*3}=\frac{-18-24\sqrt{2}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+24\sqrt{2}}{2*3}=\frac{-18+24\sqrt{2}}{6}
$