4b(3b-2)=10b+6

Solution for 4b(3b-2)=10b+6 equation:

4b(3b-2)=10b+6

We move all terms to the left:

4b(3b-2)-(10b+6)=0

We multiply parentheses

12b^2-8b-(10b+6)=0

We get rid of parentheses

12b^2-8b-10b-6=0

We add all the numbers together, and all the variables

12b^2-18b-6=0

a = 12; b = -18; c = -6;
Δ = b2-4ac
Δ = -182-4·12·(-6)
Δ = 612
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:

$\sqrt{\Delta}=\sqrt{612}=\sqrt{36*17}=\sqrt{36}*\sqrt{17}=6\sqrt{17}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{17}}{2*12}=\frac{18-6\sqrt{17}}{24}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{17}}{2*12}=\frac{18+6\sqrt{17}}{24}
$