-5(b+2)3b=-2(1+5b)+6

Solution for -5(b+2)3b=-2(1+5b)+6 equation:

-5(b+2)3b=-2(1+5b)+6

We move all terms to the left:

-5(b+2)3b-(-2(1+5b)+6)=0

We add all the numbers together, and all the variables

-5(b+2)3b-(-2(5b+1)+6)=0

We multiply parentheses

-15b^2-30b-(-2(5b+1)+6)=0


We calculate terms in parentheses: -(-2(5b+1)+6), so:

-2(5b+1)+6

We multiply parentheses

-10b-2+6

We add all the numbers together, and all the variables

-10b+4

Back to the equation:

-(-10b+4)


We get rid of parentheses

-15b^2-30b+10b-4=0

We add all the numbers together, and all the variables

-15b^2-20b-4=0

a = -15; b = -20; c = -4;
Δ = b2-4ac
Δ = -202-4·(-15)·(-4)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{10}}{2*-15}=\frac{20-4\sqrt{10}}{-30}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{10}}{2*-15}=\frac{20+4\sqrt{10}}{-30}
$