b(b+9)=4(5+2b)=

Solution for b(b+9)=4(5+2b)= equation:

b(b+9)=4(5+2b)=

We move all terms to the left:

b(b+9)-(4(5+2b))=0

We add all the numbers together, and all the variables

b(b+9)-(4(2b+5))=0

We multiply parentheses

b^2+9b-(4(2b+5))=0


We calculate terms in parentheses: -(4(2b+5)), so:

4(2b+5)

We multiply parentheses

8b+20

Back to the equation:

-(8b+20)


We get rid of parentheses

b^2+9b-8b-20=0

We add all the numbers together, and all the variables

b^2+b-20=0

a = 1; b = 1; c = -20;
Δ = b2-4ac
Δ = 12-4·1·(-20)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*1}=\frac{-10}{2}
=-5
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*1}=\frac{8}{2}
=4
$