3x^2-18x+23=0

Solution for 3x^2-18x+23=0 equation:

3x^2-18x+23=0

a = 3; b = -18; c = +23;
Δ = b2-4ac
Δ = -182-4·3·23
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-4\sqrt{3}}{2*3}=\frac{18-4\sqrt{3}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+4\sqrt{3}}{2*3}=\frac{18+4\sqrt{3}}{6}
$