What are the solutions to the equation 8f^2+11f=2f^2+3(2f-5)+21 ?

The solutions to the equation 8f^2+11f=2f^2+3(2f-5)+21 are f=-3/2 ,f= 2/3

Find the zeros of 8f^2+11f=2f^2+3(2f-5)+21

The zeros of 8f^2+11f=2f^2+3(2f-5)+21 are f=-3/2 ,f= 2/3

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8 f^{2} + 11 f = 2 f^{2} + 3 (2 f - 5) + 21

f = - \frac{3}{2}, f = \frac{2}{3}

Solution steps

8 f^{2} + 11 f = 2 f^{2} + 3 (2 f - 5) + 21

Expand

2 f^{2} + 3 (2 f - 5) + 21 : 2 f^{2} + 6 f + 6

8 f^{2} + 11 f = 2 f^{2} + 6 f + 6

Switch sides

2 f^{2} + 6 f + 6 = 8 f^{2} + 11 f

Move

11 f

to the left side

2 f^{2} - 5 f + 6 = 8 f^{2}

Move

8 f^{2}

to the left side

- 6 f^{2} - 5 f + 6 = 0

Solve with the quadratic formula

f_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 6 ) \cdot 6}{2 ( - 6 )}

(- 5)^{2} - 4 (- 6) \cdot 6 = 13

f_{1, 2} = \frac{- ( - 5 ) \pm 13}{2 ( - 6 )}

Separate the solutions

f_{1} = \frac{- ( - 5 ) + 13}{2 ( - 6 )}, f_{2} = \frac{- ( - 5 ) - 13}{2 ( - 6 )}

f = \frac{- ( - 5 ) + 13}{2 ( - 6 )} : - \frac{3}{2}

f = \frac{- ( - 5 ) - 13}{2 ( - 6 )} : \frac{2}{3}

The solutions to the quadratic equation are:

f = - \frac{3}{2}, f = \frac{2}{3}

y, 3 y - 2 x y - 3 y^{2} = 0

(2 x - 1) (x + 3) = - 2 x

a x^{2} + 2 x + a = 5 x^{2} - 3 a x + 4

(36 - x)^{2} + (18 - x)^{2} = x^{2}

- 0.02 x^{2} + x + 2.6 = 0

What are the solutions to the equation 8f^2+11f=2f^2+3(2f-5)+21 ?
The solutions to the equation 8f^2+11f=2f^2+3(2f-5)+21 are f=-3/2 ,f= 2/3
Find the zeros of 8f^2+11f=2f^2+3(2f-5)+21
The zeros of 8f^2+11f=2f^2+3(2f-5)+21 are f=-3/2 ,f= 2/3