What are the solutions to the equation 200-4x=100+0.2x^2 ?

The solutions to the equation 200-4x=100+0.2x^2 are x=10(sqrt(6)-1),x=-10(1+sqrt(6))

Find the zeros of 200-4x=100+0.2x^2

The zeros of 200-4x=100+0.2x^2 are x=10(sqrt(6)-1),x=-10(1+sqrt(6))

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200 - 4 x = 100 + 0.2 x^{2}

x = 10 (6 - 1), x = - 10 (1 + 6)

Decimal Notation

x = 14.49489 \dots, x = - 34.49489 \dots

Solution steps

200 - 4 x = 100 + 0.2 x^{2}

Multiply both sides by

10

2000 - 40 x = 1000 + 2 x^{2}

Switch sides

1000 + 2 x^{2} = 2000 - 40 x

Move

40 x

to the left side

1000 + 2 x^{2} + 40 x = 2000

Move

2000

to the left side

2 x^{2} + 40 x - 1000 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 \cdot 2 ( - 1000 )}{2 \cdot 2}

4 0^{2} - 4 \cdot 2 (- 1000) = 406

x_{1, 2} = \frac{- 40 \pm 40 6}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 40 + 40 6}{2 \cdot 2}, x_{2} = \frac{- 40 - 40 6}{2 \cdot 2}

x = \frac{- 40 + 40 6}{2 \cdot 2} : 10 (6 - 1)

x = \frac{- 40 - 40 6}{2 \cdot 2} : - 10 (1 + 6)

The solutions to the quadratic equation are:

x = 10 (6 - 1), x = - 10 (1 + 6)

x^{2} + 2 x = 5

25 x^{2} - 169 = 0

7 w^{2} + 2 w - 3 = 0

- 2 n^{2} - 5 = - 103

x, 4 x^{2} - 3 y^{2} + 12 z^{2} + 12 = 0

What are the solutions to the equation 200-4x=100+0.2x^2 ?
The solutions to the equation 200-4x=100+0.2x^2 are x=10(sqrt(6)-1),x=-10(1+sqrt(6))
Find the zeros of 200-4x=100+0.2x^2
The zeros of 200-4x=100+0.2x^2 are x=10(sqrt(6)-1),x=-10(1+sqrt(6))