5(b+3)-5(b2-1)=b2+7(3-b)-1

Solution for 5(b+3)-5(b2-1)=b2+7(3-b)-1 equation:

5(b+3)-5(b2-1)=b2+7(3-b)-1

We move all terms to the left:

5(b+3)-5(b2-1)-(b2+7(3-b)-1)=0

We add all the numbers together, and all the variables

-5(+b^2-1)+5(b+3)-(b2+7(-1b+3)-1)=0

We multiply parentheses

-5b^2+5b-(b2+7(-1b+3)-1)+5+15=0


We calculate terms in parentheses: -(b2+7(-1b+3)-1), so:

b2+7(-1b+3)-1

We add all the numbers together, and all the variables

b^2+7(-1b+3)-1

We multiply parentheses

b^2-7b+21-1

We add all the numbers together, and all the variables

b^2-7b+20

Back to the equation:

-(b^2-7b+20)


We add all the numbers together, and all the variables

-5b^2+5b-(b^2-7b+20)+20=0

We get rid of parentheses

-5b^2-b^2+5b+7b-20+20=0

We add all the numbers together, and all the variables

-6b^2+12b=0

a = -6; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·(-6)·0
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{-(12)-12}{2*-6}=\frac{-24}{-12}
=+2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*-6}=\frac{0}{-12}
=0
$