b(b+9)=4(5+3b)

Solution for b(b+9)=4(5+3b) equation:

b(b+9)=4(5+3b)

We move all terms to the left:

b(b+9)-(4(5+3b))=0

We add all the numbers together, and all the variables

b(b+9)-(4(3b+5))=0

We multiply parentheses

b^2+9b-(4(3b+5))=0


We calculate terms in parentheses: -(4(3b+5)), so:

4(3b+5)

We multiply parentheses

12b+20

Back to the equation:

-(12b+20)


We get rid of parentheses

b^2+9b-12b-20=0

We add all the numbers together, and all the variables

b^2-3b-20=0

a = 1; b = -3; c = -20;
Δ = b2-4ac
Δ = -32-4·1·(-20)
Δ = 89
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{89}}{2*1}=\frac{3-\sqrt{89}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{89}}{2*1}=\frac{3+\sqrt{89}}{2}
$