b^2+5b=35+3b

Solution for b^2+5b=35+3b equation:

b^2+5b=35+3b

We move all terms to the left:

b^2+5b-(35+3b)=0

We add all the numbers together, and all the variables

b^2+5b-(3b+35)=0

We get rid of parentheses

b^2+5b-3b-35=0

We add all the numbers together, and all the variables

b^2+2b-35=0

a = 1; b = 2; c = -35;
Δ = b2-4ac
Δ = 22-4·1·(-35)
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{-(2)-12}{2*1}=\frac{-14}{2}
=-7
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+12}{2*1}=\frac{10}{2}
=5
$