3x^2+9x+0.5=0

Solution for 3x^2+9x+0.5=0 equation:

3x^2+9x+0.5=0

a = 3; b = 9; c = +0.5;
Δ = b2-4ac
Δ = 92-4·3·0.5
Δ = 75
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:

$\sqrt{\Delta}=\sqrt{75}=\sqrt{25*3}=\sqrt{25}*\sqrt{3}=5\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-5\sqrt{3}}{2*3}=\frac{-9-5\sqrt{3}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+5\sqrt{3}}{2*3}=\frac{-9+5\sqrt{3}}{6}
$