15(b-2)+15=b(2b-9)

Solution for 15(b-2)+15=b(2b-9) equation:

15(b-2)+15=b(2b-9)

We move all terms to the left:

15(b-2)+15-(b(2b-9))=0

We multiply parentheses

15b-(b(2b-9))-30+15=0


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

b(2b-9)

We multiply parentheses

2b^2-9b

Back to the equation:

-(2b^2-9b)


We add all the numbers together, and all the variables

15b-(2b^2-9b)-15=0

We get rid of parentheses

-2b^2+15b+9b-15=0

We add all the numbers together, and all the variables

-2b^2+24b-15=0

a = -2; b = 24; c = -15;
Δ = b2-4ac
Δ = 242-4·(-2)·(-15)
Δ = 456
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{456}=\sqrt{4*114}=\sqrt{4}*\sqrt{114}=2\sqrt{114}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-2\sqrt{114}}{2*-2}=\frac{-24-2\sqrt{114}}{-4}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+2\sqrt{114}}{2*-2}=\frac{-24+2\sqrt{114}}{-4}
$