3x^2+6x-4.5=0

Solution for 3x^2+6x-4.5=0 equation:

3x^2+6x-4.5=0

a = 3; b = 6; c = -4.5;
Δ = b2-4ac
Δ = 62-4·3·(-4.5)
Δ = 90
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:

$\sqrt{\Delta}=\sqrt{90}=\sqrt{9*10}=\sqrt{9}*\sqrt{10}=3\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-3\sqrt{10}}{2*3}=\frac{-6-3\sqrt{10}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+3\sqrt{10}}{2*3}=\frac{-6+3\sqrt{10}}{6}
$