9r^2-4r-4=0

Solution for 9r^2-4r-4=0 equation:

9r^2-4r-4=0

a = 9; b = -4; c = -4;
Δ = b2-4ac
Δ = -42-4·9·(-4)
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{-(-4)-4\sqrt{10}}{2*9}=\frac{4-4\sqrt{10}}{18}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{10}}{2*9}=\frac{4+4\sqrt{10}}{18}
$