5b^2+32=-28b

Solution for 5b^2+32=-28b equation:

5b^2+32=-28b

We move all terms to the left:

5b^2+32-(-28b)=0

We get rid of parentheses

5b^2+28b+32=0

a = 5; b = 28; c = +32;
Δ = b2-4ac
Δ = 282-4·5·32
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-12}{2*5}=\frac{-40}{10}
=-4
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+12}{2*5}=\frac{-16}{10}
=-1+3/5
$