11r^2-13=4r

Solution for 11r^2-13=4r equation:

11r^2-13=4r

We move all terms to the left:

11r^2-13-(4r)=0

a = 11; b = -4; c = -13;
Δ = b2-4ac
Δ = -42-4·11·(-13)
Δ = 588
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:

$\sqrt{\Delta}=\sqrt{588}=\sqrt{196*3}=\sqrt{196}*\sqrt{3}=14\sqrt{3}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-14\sqrt{3}}{2*11}=\frac{4-14\sqrt{3}}{22}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+14\sqrt{3}}{2*11}=\frac{4+14\sqrt{3}}{22}
$