4(r+2)r=6

Solution for 4(r+2)r=6 equation:

4(r+2)r=6

We move all terms to the left:

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

We multiply parentheses

4r^2+8r-6=0

a = 4; b = 8; c = -6;
Δ = b2-4ac
Δ = 82-4·4·(-6)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{10}}{2*4}=\frac{-8-4\sqrt{10}}{8}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{10}}{2*4}=\frac{-8+4\sqrt{10}}{8}
$