What are the solutions to the equation 25x^2+64=289 ?

The solutions to the equation 25x^2+64=289 are x=3,x=-3

Find the zeros of 25x^2+64=289

The zeros of 25x^2+64=289 are x=3,x=-3

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25 x^{2} + 64 = 289

x = 3, x = - 3

Solution steps

25 x^{2} + 64 = 289

Move

64

to the right side

25 x^{2} = 225

Divide both sides by

25

x^{2} = 9

For

x^{2} = f (a)

the solutions are

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

x = 9, x = - 9

9 = 3

- 9 = - 3

x = 3, x = - 3

(2 a^{3} b^{4})^{\frac{3}{2}}

w^{2} + w - 12

\frac{4}{5} x - 8 = 12

(2 x - 7)^{2} + 8 = 0

x^{2} - 12 x + 27 < 0

What are the solutions to the equation 25x^2+64=289 ?
The solutions to the equation 25x^2+64=289 are x=3,x=-3
Find the zeros of 25x^2+64=289
The zeros of 25x^2+64=289 are x=3,x=-3