8x^2+12=-28x

Solution for 8x^2+12=-28x equation:

8x^2+12=-28x

We move all terms to the left:

8x^2+12-(-28x)=0

We get rid of parentheses

8x^2+28x+12=0

a = 8; b = 28; c = +12;
Δ = b2-4ac
Δ = 282-4·8·12
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-20}{2*8}=\frac{-48}{16}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+20}{2*8}=\frac{-8}{16}
=-1/2
$