3x^2+27x+4.5=0

Solution for 3x^2+27x+4.5=0 equation:

3x^2+27x+4.5=0

a = 3; b = 27; c = +4.5;
Δ = b2-4ac
Δ = 272-4·3·4.5
Δ = 675
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:

$\sqrt{\Delta}=\sqrt{675}=\sqrt{225*3}=\sqrt{225}*\sqrt{3}=15\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-15\sqrt{3}}{2*3}=\frac{-27-15\sqrt{3}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+15\sqrt{3}}{2*3}=\frac{-27+15\sqrt{3}}{6}
$