6x2-42x+12=0

Solution for 6x2-42x+12=0 equation:

6x^2-42x+12=0

a = 6; b = -42; c = +12;
Δ = b2-4ac
Δ = -422-4·6·12
Δ = 1476
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:

$\sqrt{\Delta}=\sqrt{1476}=\sqrt{36*41}=\sqrt{36}*\sqrt{41}=6\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-6\sqrt{41}}{2*6}=\frac{42-6\sqrt{41}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+6\sqrt{41}}{2*6}=\frac{42+6\sqrt{41}}{12}
$