3000r^2+2000r^2=408

Solution for 3000r^2+2000r^2=408 equation:

3000r^2+2000r^2=408

We move all terms to the left:

3000r^2+2000r^2-(408)=0

We add all the numbers together, and all the variables

5000r^2-408=0

a = 5000; b = 0; c = -408;
Δ = b2-4ac
Δ = 02-4·5000·(-408)
Δ = 8160000
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:

$\sqrt{\Delta}=\sqrt{8160000}=\sqrt{160000*51}=\sqrt{160000}*\sqrt{51}=400\sqrt{51}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-400\sqrt{51}}{2*5000}=\frac{0-400\sqrt{51}}{10000}
=-\frac{400\sqrt{51}}{10000}
=-\frac{\sqrt{51}}{25}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+400\sqrt{51}}{2*5000}=\frac{0+400\sqrt{51}}{10000}
=\frac{400\sqrt{51}}{10000}
=\frac{\sqrt{51}}{25}
$