12x^2 + 20x + 28 solve for x

Answer 1

You don't show an equation to solve.
Did you mean,
solve 12x^2 + 20x + 28 = 0 ?

if so, I would first divide every term by 4
3x^2 + 5x + 7 = 0
Now use the quadratic formula
x = (-5 ± √(25-4(3)(7))/6
= (-5 ± √-59)/6

giving us two complex roots.