28=2b(b-2)-4b

Solution for 28=2b(b-2)-4b equation:

28=2b(b-2)-4b

We move all terms to the left:

28-(2b(b-2)-4b)=0


We calculate terms in parentheses: -(2b(b-2)-4b), so:

2b(b-2)-4b

We add all the numbers together, and all the variables

-4b+2b(b-2)

We multiply parentheses

2b^2-4b-4b

We add all the numbers together, and all the variables

2b^2-8b

Back to the equation:

-(2b^2-8b)


We get rid of parentheses

-2b^2+8b+28=0

a = -2; b = 8; c = +28;
Δ = b2-4ac
Δ = 82-4·(-2)·28
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-12\sqrt{2}}{2*-2}=\frac{-8-12\sqrt{2}}{-4}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+12\sqrt{2}}{2*-2}=\frac{-8+12\sqrt{2}}{-4}
$