What are the solutions to the equation -1/4 x^2+4x=-2 ?

The solutions to the equation -1/4 x^2+4x=-2 are x=-2(3sqrt(2)-4),x=2(4+3sqrt(2))

Find the zeros of -1/4 x^2+4x=-2

The zeros of -1/4 x^2+4x=-2 are x=-2(3sqrt(2)-4),x=2(4+3sqrt(2))

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- \frac{1}{4} x^{2} + 4 x = - 2

x = - 2 (32 - 4), x = 2 (4 + 32)

Decimal Notation

x = - 0.48528 \dots, x = 16.48528 \dots

Solution steps

- \frac{1}{4} x^{2} + 4 x = - 2

Multiply both sides by

4

- x^{2} + 16 x = - 8

Move

8

to the left side

- x^{2} + 16 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 ( - 1 ) \cdot 8}{2 ( - 1 )}

1 6^{2} - 4 (- 1) \cdot 8 = 122

x_{1, 2} = \frac{- 16 \pm 12 2}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 16 + 12 2}{2 ( - 1 )}, x_{2} = \frac{- 16 - 12 2}{2 ( - 1 )}

x = \frac{- 16 + 12 2}{2 ( - 1 )} : - 2 (32 - 4)

x = \frac{- 16 - 12 2}{2 ( - 1 )} : 2 (4 + 32)

The solutions to the quadratic equation are:

x = - 2 (32 - 4), x = 2 (4 + 32)

\frac{1}{10} q^{2} + 5 = 20 - \frac{q}{2}

x, x^{2} + y^{2} + z^{2} = 6 x - 4 y - 10 z

x^{2} - 3 x - 4 = 0

4.9 t^{2} + 8 t - 25 = 0

x, z^{2} = 4 + 4 (x^{2} + y^{2})

What are the solutions to the equation -1/4 x^2+4x=-2 ?
The solutions to the equation -1/4 x^2+4x=-2 are x=-2(3sqrt(2)-4),x=2(4+3sqrt(2))
Find the zeros of -1/4 x^2+4x=-2
The zeros of -1/4 x^2+4x=-2 are x=-2(3sqrt(2)-4),x=2(4+3sqrt(2))