Solve quadratic?

You have to make it look like a quadratic equation, the equation you gave is actually a quartic.

Therefore, do as follows:

Let r² = y which implies that r^4 = y²

Now subsitute into your original equation, and it will be a quadratic euqation!

Therefore your equation becomes:

0.21r^4 - 2.42 r^2 + 1.21 = 0
0.21y² - 2.42y + 1.21 = 0; after substituting r² = y and r^4 = y²
21y² - 242y + 121 = 0; I multiplied the above equation by 100
(21y - 11)(y - 11) = 0; after factoring

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Now solve:

21y - 11 = 0
y = 11/21

But r² = y

Therefore,

r² = 11/21
r = ± √(11/21)

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Now solve

y - 11 = 0
y = 11

But remember r² = y

Therefore,

r² = 11
r = ±√11

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Hope this helps!

Multiplying Equation by 100 and supposing r square as x, we get

21x^2-242x+121=0

Solve to get x (Use x= [-b+/-sq rt. (b^2 - 4ac)]/2a
Find r by taking square root of x

0.21r^4 - 2.42r^2 + 1.21 = 0
21r^4 - 242r^2 + 121 = 0
(21r^2 - 11)(r^2 - 11) = 0

21r^2 - 11 = 0
21r^2 = 11
r^2 = 11/21
r = ±√(11/21)

r^2 - 11 = 0
r^2 = 11
r = ±√11

∴ r = ±√(11/21) , ±√11

For this problem, you'll need a graphics calculator or manullay plot the function. I used the range [-5,5] and found 3 roots. You may try the attached source should you wish.
The 3 roots are,
r=-3.3166
-0.7237
0.7237
This is not a quadratic; it is a 4th degree polynomial.