What are the solutions to the equation -4x=-2x^2+16 ?

The solutions to the equation -4x=-2x^2+16 are x=-2,x=4

Find the zeros of -4x=-2x^2+16

The zeros of -4x=-2x^2+16 are x=-2,x=4

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- 4 x = - 2 x^{2} + 16

x = - 2, x = 4

Solution steps

- 4 x = - 2 x^{2} + 16

Switch sides

- 2 x^{2} + 16 = - 4 x

Move

4 x

to the left side

- 2 x^{2} + 16 + 4 x = 0

Write in the standard form

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

- 2 x^{2} + 4 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 2 ) \cdot 16}{2 ( - 2 )}

4^{2} - 4 (- 2) \cdot 16 = 12

x_{1, 2} = \frac{- 4 \pm 12}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 4 + 12}{2 ( - 2 )}, x_{2} = \frac{- 4 - 12}{2 ( - 2 )}

x = \frac{- 4 + 12}{2 ( - 2 )} : - 2

x = \frac{- 4 - 12}{2 ( - 2 )} : 4

The solutions to the quadratic equation are:

x = - 2, x = 4

t, s (t) = 4.9 t^{2} + 88.2 t + C

x^{2} - 2 π x - 2 π = 0

0 = - 3 x^{2} + 13 x + 10

x^{2} - 8 x + 4 = 0

c^{2} = 4^{2} + 8^{2}

What are the solutions to the equation -4x=-2x^2+16 ?
The solutions to the equation -4x=-2x^2+16 are x=-2,x=4
Find the zeros of -4x=-2x^2+16
The zeros of -4x=-2x^2+16 are x=-2,x=4