5(b+7)=3(b2+9)

Solution for 5(b+7)=3(b2+9) equation:

5(b+7)=3(b2+9)

We move all terms to the left:

5(b+7)-(3(b2+9))=0

We add all the numbers together, and all the variables

-(3(+b^2+9))+5(b+7)=0

We multiply parentheses

-(3(+b^2+9))+5b+35=0


We calculate terms in parentheses: -(3(+b^2+9)), so:

3(+b^2+9)

We multiply parentheses

3b^2+27

Back to the equation:

-(3b^2+27)


We add all the numbers together, and all the variables

5b-(3b^2+27)+35=0

We get rid of parentheses

-3b^2+5b-27+35=0

We add all the numbers together, and all the variables

-3b^2+5b+8=0

a = -3; b = 5; c = +8;
Δ = b2-4ac
Δ = 52-4·(-3)·8
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*-3}=\frac{-16}{-6}
=2+2/3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*-3}=\frac{6}{-6}
=-1
$