(R)(Q)=3000Q-20q2

Solution for (R)(Q)=3000Q-20q2 equation:

()()=3000-20R^2

We move all terms to the left:

()()-(3000-20R^2)=0

We add all the numbers together, and all the variables

-(3000-20R^2)=0

We get rid of parentheses

20R^2-3000=0

a = 20; b = 0; c = -3000;
Δ = b2-4ac
Δ = 02-4·20·(-3000)
Δ = 240000
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:

$\sqrt{\Delta}=\sqrt{240000}=\sqrt{40000*6}=\sqrt{40000}*\sqrt{6}=200\sqrt{6}$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-200\sqrt{6}}{2*20}=\frac{0-200\sqrt{6}}{40}
=-\frac{200\sqrt{6}}{40}
=-5\sqrt{6}
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+200\sqrt{6}}{2*20}=\frac{0+200\sqrt{6}}{40}
=\frac{200\sqrt{6}}{40}
=5\sqrt{6}
$