What are the solutions to the equation 240=-0.1x^2+400 ?

The solutions to the equation 240=-0.1x^2+400 are x=40,x=-40

Find the zeros of 240=-0.1x^2+400

The zeros of 240=-0.1x^2+400 are x=40,x=-40

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240 = - 0.1 x^{2} + 400

x = 40, x = - 40

Solution steps

240 = - 0.1 x^{2} + 400

Switch sides

- 0.1 x^{2} + 400 = 240

Move

400

to the right side

- 0.1 x^{2} = - 160

Divide both sides by

- 0.1

x^{2} = 1600

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = 1600, x = - 1600

1600 = 40

- 1600 = - 40

x = 40, x = - 40

2 = 3 t^{2} - 4

d^{2} - 17 d + 16 = 0

32 x^{2} - 5 x - 2 = 0

x^{2} - 15 x + 20 = 0

6 x^{2} + 12 x = 0

What are the solutions to the equation 240=-0.1x^2+400 ?
The solutions to the equation 240=-0.1x^2+400 are x=40,x=-40
Find the zeros of 240=-0.1x^2+400
The zeros of 240=-0.1x^2+400 are x=40,x=-40