12x^2-52x+42=0

Solution for 12x^2-52x+42=0 equation:

12x^2-52x+42=0

a = 12; b = -52; c = +42;
Δ = b2-4ac
Δ = -522-4·12·42
Δ = 688
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:

$\sqrt{\Delta}=\sqrt{688}=\sqrt{16*43}=\sqrt{16}*\sqrt{43}=4\sqrt{43}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-52)-4\sqrt{43}}{2*12}=\frac{52-4\sqrt{43}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-52)+4\sqrt{43}}{2*12}=\frac{52+4\sqrt{43}}{24}
$