4b^2-32b=-28

Solution for 4b^2-32b=-28 equation:

4b^2-32b=-28

We move all terms to the left:

4b^2-32b-(-28)=0

We add all the numbers together, and all the variables

4b^2-32b+28=0

a = 4; b = -32; c = +28;
Δ = b2-4ac
Δ = -322-4·4·28
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-24}{2*4}=\frac{8}{8}
=1
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+24}{2*4}=\frac{56}{8}
=7
$