b^2+2b=-3b+24

Solution for b^2+2b=-3b+24 equation:

b^2+2b=-3b+24

We move all terms to the left:

b^2+2b-(-3b+24)=0

We get rid of parentheses

b^2+2b+3b-24=0

We add all the numbers together, and all the variables

b^2+5b-24=0

a = 1; b = 5; c = -24;
Δ = b2-4ac
Δ = 52-4·1·(-24)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-11}{2*1}=\frac{-16}{2}
=-8
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*1}=\frac{6}{2}
=3
$