2b(3b-10)=b-5

Solution for 2b(3b-10)=b-5 equation:

2b(3b-10)=b-5

We move all terms to the left:

2b(3b-10)-(b-5)=0

We multiply parentheses

6b^2-20b-(b-5)=0

We get rid of parentheses

6b^2-20b-b+5=0

We add all the numbers together, and all the variables

6b^2-21b+5=0

a = 6; b = -21; c = +5;
Δ = b2-4ac
Δ = -212-4·6·5
Δ = 321
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{-(-21)-\sqrt{321}}{2*6}=\frac{21-\sqrt{321}}{12}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{321}}{2*6}=\frac{21+\sqrt{321}}{12}
$