-4b+6b(-6b+2)=-2b-140

Solution for -4b+6b(-6b+2)=-2b-140 equation:

-4b+6b(-6b+2)=-2b-140

We move all terms to the left:

-4b+6b(-6b+2)-(-2b-140)=0

We multiply parentheses

-36b^2-4b+12b-(-2b-140)=0

We get rid of parentheses

-36b^2-4b+12b+2b+140=0

We add all the numbers together, and all the variables

-36b^2+10b+140=0

a = -36; b = 10; c = +140;
Δ = b2-4ac
Δ = 102-4·(-36)·140
Δ = 20260
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:

$\sqrt{\Delta}=\sqrt{20260}=\sqrt{4*5065}=\sqrt{4}*\sqrt{5065}=2\sqrt{5065}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{5065}}{2*-36}=\frac{-10-2\sqrt{5065}}{-72}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{5065}}{2*-36}=\frac{-10+2\sqrt{5065}}{-72}
$