-4(r+2)=4(2-2r)r=

Solution for -4(r+2)=4(2-2r)r= equation:

-4(r+2)=4(2-2r)r=

We move all terms to the left:

-4(r+2)-(4(2-2r)r)=0

We add all the numbers together, and all the variables

-4(r+2)-(4(-2r+2)r)=0

We multiply parentheses

-4r-(4(-2r+2)r)-8=0


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

4(-2r+2)r

We multiply parentheses

-8r^2+8r

Back to the equation:

-(-8r^2+8r)


We get rid of parentheses

8r^2-8r-4r-8=0

We add all the numbers together, and all the variables

8r^2-12r-8=0

a = 8; b = -12; c = -8;
Δ = b2-4ac
Δ = -122-4·8·(-8)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-20}{2*8}=\frac{-8}{16}
=-1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+20}{2*8}=\frac{32}{16}
=2
$