(9r^2+5r-11)-(-2r^2+9r-8)=0

Solution for (9r^2+5r-11)-(-2r^2+9r-8)=0 equation:

(9r^2+5r-11)-(-2r^2+9r-8)=0

We get rid of parentheses

2r^2+9r^2-9r+5r+8-11=0

We add all the numbers together, and all the variables

11r^2-4r-3=0

a = 11; b = -4; c = -3;
Δ = b2-4ac
Δ = -42-4·11·(-3)
Δ = 148
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:

$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{37}}{2*11}=\frac{4-2\sqrt{37}}{22}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{37}}{2*11}=\frac{4+2\sqrt{37}}{22}
$