-3(1+b)=-b(b+6)

Solution for -3(1+b)=-b(b+6) equation:

-3(1+b)=-b(b+6)

We move all terms to the left:

-3(1+b)-(-b(b+6))=0

We add all the numbers together, and all the variables

-3(b+1)-(-b(b+6))=0

We multiply parentheses

-3b-(-b(b+6))-3=0


We calculate terms in parentheses: -(-b(b+6)), so:

-b(b+6)

We multiply parentheses

-b^2-6b

We add all the numbers together, and all the variables

-1b^2-6b

Back to the equation:

-(-1b^2-6b)


We get rid of parentheses

1b^2+6b-3b-3=0

We add all the numbers together, and all the variables

b^2+3b-3=0

a = 1; b = 3; c = -3;
Δ = b2-4ac
Δ = 32-4·1·(-3)
Δ = 21
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{21}}{2*1}=\frac{-3-\sqrt{21}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{21}}{2*1}=\frac{-3+\sqrt{21}}{2}
$