10b(b+3)=11b+3

Solution for 10b(b+3)=11b+3 equation:

10b(b+3)=11b+3

We move all terms to the left:

10b(b+3)-(11b+3)=0

We multiply parentheses

10b^2+30b-(11b+3)=0

We get rid of parentheses

10b^2+30b-11b-3=0

We add all the numbers together, and all the variables

10b^2+19b-3=0

a = 10; b = 19; c = -3;
Δ = b2-4ac
Δ = 192-4·10·(-3)
Δ = 481
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{-(19)-\sqrt{481}}{2*10}=\frac{-19-\sqrt{481}}{20}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{481}}{2*10}=\frac{-19+\sqrt{481}}{20}
$