What are the solutions to the equation 9=39-4.9(x)^2 ?

The solutions to the equation 9=39-4.9(x)^2 are x=(10sqrt(3))/7 ,x=-(10sqrt(3))/7

Find the zeros of 9=39-4.9(x)^2

The zeros of 9=39-4.9(x)^2 are x=(10sqrt(3))/7 ,x=-(10sqrt(3))/7

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9 = 39 - 4.9 (x)^{2}

x = \frac{10 3}{7}, x = - \frac{10 3}{7}

Solution steps

9 = 39 - 4.9 x^{2}

Switch sides

39 - 4.9 x^{2} = 9

Move

39

to the right side

- 4.9 x^{2} = - 30

Multiply both sides by

10

- 49 x^{2} = - 300

Divide both sides by

- 49

x^{2} = \frac{300}{49}

For

x^{2} = f (a)

the solutions are

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

x = \frac{300}{49}, x = - \frac{300}{49}

\frac{300}{49} = \frac{10 3}{7}

- \frac{300}{49} = - \frac{10 3}{7}

x = \frac{10 3}{7}, x = - \frac{10 3}{7}

(1 + \frac{0.05}{12})^{12} - 1

- 2 x = 38

x^{3} - 3 x^{2} - 9 x - 5

- \frac{1}{4} x^{2} + x + 1 = 0

∣ 3 z - 9 ∣ > ∣ 3 z - 4 ∣

What are the solutions to the equation 9=39-4.9(x)^2 ?
The solutions to the equation 9=39-4.9(x)^2 are x=(10sqrt(3))/7 ,x=-(10sqrt(3))/7
Find the zeros of 9=39-4.9(x)^2
The zeros of 9=39-4.9(x)^2 are x=(10sqrt(3))/7 ,x=-(10sqrt(3))/7