3x^2+27x+18=0

Solution for 3x^2+27x+18=0 equation:

3x^2+27x+18=0

a = 3; b = 27; c = +18;
Δ = b2-4ac
Δ = 272-4·3·18
Δ = 513
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:

$\sqrt{\Delta}=\sqrt{513}=\sqrt{9*57}=\sqrt{9}*\sqrt{57}=3\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-3\sqrt{57}}{2*3}=\frac{-27-3\sqrt{57}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+3\sqrt{57}}{2*3}=\frac{-27+3\sqrt{57}}{6}
$