(r+1)2r=6

Solution for (r+1)2r=6 equation:

(r+1)2r=6

We move all terms to the left:

(r+1)2r-(6)=0

We multiply parentheses

2r^2+2r-6=0

a = 2; b = 2; c = -6;
Δ = b2-4ac
Δ = 22-4·2·(-6)
Δ = 52
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{13}}{2*2}=\frac{-2-2\sqrt{13}}{4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{13}}{2*2}=\frac{-2+2\sqrt{13}}{4}
$