3b^2+28b=-32

Solution for 3b^2+28b=-32 equation:

3b^2+28b=-32

We move all terms to the left:

3b^2+28b-(-32)=0

We add all the numbers together, and all the variables

3b^2+28b+32=0

a = 3; b = 28; c = +32;
Δ = b2-4ac
Δ = 282-4·3·32
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-20}{2*3}=\frac{-48}{6}
=-8
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+20}{2*3}=\frac{-8}{6}
=-1+1/3
$