4=2(r+1)r=

Solution for 4=2(r+1)r= equation:

4=2(r+1)r=

We move all terms to the left:

4-(2(r+1)r)=0


We calculate terms in parentheses: -(2(r+1)r), so:

2(r+1)r

We multiply parentheses

2r^2+2r

Back to the equation:

-(2r^2+2r)


We get rid of parentheses

-2r^2-2r+4=0

a = -2; b = -2; c = +4;
Δ = b2-4ac
Δ = -22-4·(-2)·4
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6}{2*-2}=\frac{-4}{-4}
=1
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6}{2*-2}=\frac{8}{-4}
=-2
$