12-28x+12x^2=0

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

12-28x+12x^2=0

a = 12; b = -28; c = +12;
Δ = b2-4ac
Δ = -282-4·12·12
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{13}}{2*12}=\frac{28-4\sqrt{13}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{13}}{2*12}=\frac{28+4\sqrt{13}}{24}
$