9r^2+9r-28=0

Solution for 9r^2+9r-28=0 equation:

9r^2+9r-28=0

a = 9; b = 9; c = -28;
Δ = b2-4ac
Δ = 92-4·9·(-28)
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-33}{2*9}=\frac{-42}{18}
=-2+1/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+33}{2*9}=\frac{24}{18}
=1+1/3
$