49x^2-28x+4=12

Solution for 49x^2-28x+4=12 equation:

49x^2-28x+4=12

We move all terms to the left:

49x^2-28x+4-(12)=0

We add all the numbers together, and all the variables

49x^2-28x-8=0

a = 49; b = -28; c = -8;
Δ = b2-4ac
Δ = -282-4·49·(-8)
Δ = 2352
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:

$\sqrt{\Delta}=\sqrt{2352}=\sqrt{784*3}=\sqrt{784}*\sqrt{3}=28\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-28\sqrt{3}}{2*49}=\frac{28-28\sqrt{3}}{98}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+28\sqrt{3}}{2*49}=\frac{28+28\sqrt{3}}{98}
$