What are the solutions to the equation 4x^2-9=2x^2+41 ?

The solutions to the equation 4x^2-9=2x^2+41 are x=5,x=-5

Find the zeros of 4x^2-9=2x^2+41

The zeros of 4x^2-9=2x^2+41 are x=5,x=-5

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4 x^{2} - 9 = 2 x^{2} + 41

x = 5, x = - 5

Solution steps

4 x^{2} - 9 = 2 x^{2} + 41

Move

9

to the right side

4 x^{2} = 2 x^{2} + 50

Move

2 x^{2}

to the left side

2 x^{2} = 50

Divide both sides by

2

x^{2} = 25

For

x^{2} = f (a)

the solutions are

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

x = 25, x = - 25

25 = 5

- 25 = - 5

x = 5, x = - 5

- 4 x^{2} + 16 x + 5 = 0

5 u^{2} + u = 4

z^{2} + 6 z = r^{2} - 9

X^{2} = 4^{2} + 1 6^{2}

s^{2} - 14 s + 53

What are the solutions to the equation 4x^2-9=2x^2+41 ?
The solutions to the equation 4x^2-9=2x^2+41 are x=5,x=-5
Find the zeros of 4x^2-9=2x^2+41
The zeros of 4x^2-9=2x^2+41 are x=5,x=-5