11r^2+8r-6=4r^2

Solution for 11r^2+8r-6=4r^2 equation:

11r^2+8r-6=4r^2

We move all terms to the left:

11r^2+8r-6-(4r^2)=0

determiningTheFunctionDomain
11r^2-4r^2+8r-6=0

We add all the numbers together, and all the variables

7r^2+8r-6=0

a = 7; b = 8; c = -6;
Δ = b2-4ac
Δ = 82-4·7·(-6)
Δ = 232
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:

$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{58}}{2*7}=\frac{-8-2\sqrt{58}}{14}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{58}}{2*7}=\frac{-8+2\sqrt{58}}{14}
$