What are the solutions to the equation 2r=5-4r^2 ?

The solutions to the equation 2r=5-4r^2 are r=-(1+sqrt(21))/4 ,r=(sqrt(21)-1)/4

Find the zeros of 2r=5-4r^2

The zeros of 2r=5-4r^2 are r=-(1+sqrt(21))/4 ,r=(sqrt(21)-1)/4

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2 r = 5 - 4 r^{2}

r = - \frac{1 + 21}{4}, r = \frac{21 - 1}{4}

Solution steps

2 r = 5 - 4 r^{2}

Switch sides

5 - 4 r^{2} = 2 r

Move

2 r

to the left side

5 - 4 r^{2} - 2 r = 0

Write in the standard form

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

- 4 r^{2} - 2 r + 5 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 4 ) \cdot 5}{2 ( - 4 )}

(- 2)^{2} - 4 (- 4) \cdot 5 = 221

r_{1, 2} = \frac{- ( - 2 ) \pm 2 21}{2 ( - 4 )}

Separate the solutions

r_{1} = \frac{- ( - 2 ) + 2 21}{2 ( - 4 )}, r_{2} = \frac{- ( - 2 ) - 2 21}{2 ( - 4 )}

r = \frac{- ( - 2 ) + 2 21}{2 ( - 4 )} : - \frac{1 + 21}{4}

r = \frac{- ( - 2 ) - 2 21}{2 ( - 4 )} : \frac{21 - 1}{4}

The solutions to the quadratic equation are:

r = - \frac{1 + 21}{4}, r = \frac{21 - 1}{4}

6 - 2 x < 11

x^{2} - 12 x + 27 = 0

- 3 = - 5 y

\frac{1}{\frac{π}{2}}

b + 9 = 16

What are the solutions to the equation 2r=5-4r^2 ?
The solutions to the equation 2r=5-4r^2 are r=-(1+sqrt(21))/4 ,r=(sqrt(21)-1)/4
Find the zeros of 2r=5-4r^2
The zeros of 2r=5-4r^2 are r=-(1+sqrt(21))/4 ,r=(sqrt(21)-1)/4