2X^2-28x+12=0

Solution for 2X^2-28x+12=0 equation:

2X^2-28X+12=0

a = 2; b = -28; c = +12;
Δ = b2-4ac
Δ = -282-4·2·12
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{-(-28)-4\sqrt{43}}{2*2}=\frac{28-4\sqrt{43}}{4}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{43}}{2*2}=\frac{28+4\sqrt{43}}{4}
$