What are the solutions to the equation 25-r^2=4r ?

The solutions to the equation 25-r^2=4r are r=-2-sqrt(29),r=sqrt(29)-2

Find the zeros of 25-r^2=4r

The zeros of 25-r^2=4r are r=-2-sqrt(29),r=sqrt(29)-2

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25 - r^{2} = 4 r

r = - 2 - 29, r = 29 - 2

Solution steps

25 - r^{2} = 4 r

Move

4 r

to the left side

25 - r^{2} - 4 r = 0

Write in the standard form

a x^{2} + b x + c = 0

- r^{2} - 4 r + 25 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 1 ) \cdot 25}{2 ( - 1 )}

(- 4)^{2} - 4 (- 1) \cdot 25 = 229

r_{1, 2} = \frac{- ( - 4 ) \pm 2 29}{2 ( - 1 )}

Separate the solutions

r_{1} = \frac{- ( - 4 ) + 2 29}{2 ( - 1 )}, r_{2} = \frac{- ( - 4 ) - 2 29}{2 ( - 1 )}

r = \frac{- ( - 4 ) + 2 29}{2 ( - 1 )} : - 2 - 29

r = \frac{- ( - 4 ) - 2 29}{2 ( - 1 )} : 29 - 2

The solutions to the quadratic equation are:

r = - 2 - 29, r = 29 - 2

x^{2} + 4 = 2 x^{2} - 12

3 x^{2} - 3 = 45

x^{2} - 2 = 10

5 x^{2} - 8 = 12 x

x^{2} - 6 x - 22 = 0

What are the solutions to the equation 25-r^2=4r ?
The solutions to the equation 25-r^2=4r are r=-2-sqrt(29),r=sqrt(29)-2
Find the zeros of 25-r^2=4r
The zeros of 25-r^2=4r are r=-2-sqrt(29),r=sqrt(29)-2