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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

2b^2-2b-20b-(2(5b+10))=0


We calculate terms in parentheses: -(2(5b+10)), so:

2(5b+10)

We multiply parentheses

10b+20

Back to the equation:

-(10b+20)


We add all the numbers together, and all the variables

2b^2-22b-(10b+20)=0

We get rid of parentheses

2b^2-22b-10b-20=0

We add all the numbers together, and all the variables

2b^2-32b-20=0

a = 2; b = -32; c = -20;
Δ = b2-4ac
Δ = -322-4·2·(-20)
Δ = 1184
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:

$\sqrt{\Delta}=\sqrt{1184}=\sqrt{16*74}=\sqrt{16}*\sqrt{74}=4\sqrt{74}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{74}}{2*2}=\frac{32-4\sqrt{74}}{4}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{74}}{2*2}=\frac{32+4\sqrt{74}}{4}
$