1,5x^2-12x+22=0

Solution for 1,5x^2-12x+22=0 equation:

1.5x^2-12x+22=0

a = 1.5; b = -12; c = +22;
Δ = b2-4ac
Δ = -122-4·1.5·22
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{3}}{2*1.5}=\frac{12-2\sqrt{3}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{3}}{2*1.5}=\frac{12+2\sqrt{3}}{3}
$