(3b-5)(5b+5)=(2b-1)

Solution for (3b-5)(5b+5)=(2b-1) equation:

(3b-5)(5b+5)=(2b-1)

We move all terms to the left:

(3b-5)(5b+5)-((2b-1))=0

We multiply parentheses ..

(+15b^2+15b-25b-25)-((2b-1))=0


We calculate terms in parentheses: -((2b-1)), so:

(2b-1)

We get rid of parentheses

2b-1

Back to the equation:

-(2b-1)


We get rid of parentheses

15b^2+15b-25b-2b-25+1=0

We add all the numbers together, and all the variables

15b^2-12b-24=0

a = 15; b = -12; c = -24;
Δ = b2-4ac
Δ = -122-4·15·(-24)
Δ = 1584
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:

$\sqrt{\Delta}=\sqrt{1584}=\sqrt{144*11}=\sqrt{144}*\sqrt{11}=12\sqrt{11}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12\sqrt{11}}{2*15}=\frac{12-12\sqrt{11}}{30}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12\sqrt{11}}{2*15}=\frac{12+12\sqrt{11}}{30}
$