3x^2-42x+138=0

Solution for 3x^2-42x+138=0 equation:

3x^2-42x+138=0

a = 3; b = -42; c = +138;
Δ = b2-4ac
Δ = -422-4·3·138
Δ = 108
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:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-6\sqrt{3}}{2*3}=\frac{42-6\sqrt{3}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+6\sqrt{3}}{2*3}=\frac{42+6\sqrt{3}}{6}
$