What are the solutions to the equation -7f+4f^2=-2f^2-f ?

The solutions to the equation -7f+4f^2=-2f^2-f are f=1,f=0

Find the zeros of -7f+4f^2=-2f^2-f

The zeros of -7f+4f^2=-2f^2-f are f=1,f=0

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- 7 f + 4 f^{2} = - 2 f^{2} - f

f = 1, f = 0

Solution steps

- 7 f + 4 f^{2} = - 2 f^{2} - f

Move

f

to the left side

4 f^{2} - 6 f = - 2 f^{2}

Move

2 f^{2}

to the left side

6 f^{2} - 6 f = 0

Solve with the quadratic formula

f_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 6 \cdot 0}{2 \cdot 6}

(- 6)^{2} - 4 \cdot 6 \cdot 0 = 6

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

Separate the solutions

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

f = \frac{- ( - 6 ) + 6}{2 \cdot 6} : 1

f = \frac{- ( - 6 ) - 6}{2 \cdot 6} : 0

The solutions to the quadratic equation are:

f = 1, f = 0

64 = 2 u^{2}

u^{2} + 2 u - 3 = 0

(3 x - 5) 2 - (4 x + 5) (2 x - 3) = (x - 3) 2 - 2

6 x^{2} + 3 x = 1

12 x^{2} - 96 x + 140 = 0

What are the solutions to the equation -7f+4f^2=-2f^2-f ?
The solutions to the equation -7f+4f^2=-2f^2-f are f=1,f=0
Find the zeros of -7f+4f^2=-2f^2-f
The zeros of -7f+4f^2=-2f^2-f are f=1,f=0