3x^2-20x+18=0

Solution for 3x^2-20x+18=0 equation:

3x^2-20x+18=0

a = 3; b = -20; c = +18;
Δ = b2-4ac
Δ = -202-4·3·18
Δ = 184
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:

$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{46}}{2*3}=\frac{20-2\sqrt{46}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{46}}{2*3}=\frac{20+2\sqrt{46}}{6}
$