What are the solutions to the equation 18x^2+5=4 ?

The solutions to the equation 18x^2+5=4 are x= 1/(3sqrt(2))i,x=-1/(3sqrt(2))i

Find the zeros of 18x^2+5=4

The zeros of 18x^2+5=4 are x= 1/(3sqrt(2))i,x=-1/(3sqrt(2))i

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18 x^{2} + 5 = 4

x = \frac{1}{3 2} i, x = - \frac{1}{3 2} i

Solution steps

18 x^{2} + 5 = 4

Move

5

to the right side

18 x^{2} = - 1

Divide both sides by

18

x^{2} = - \frac{1}{18}

For

x^{2} = f (a)

the solutions are

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

x = - \frac{1}{18}, x = - - \frac{1}{18}

- \frac{1}{18} = \frac{1}{3 2} i

- - \frac{1}{18} = - \frac{1}{3 2} i

x = \frac{1}{3 2} i, x = - \frac{1}{3 2} i

- 10 b = - 150

(x + 2) (x - 5) < 0

1. 5^{\frac{1}{10}}

\frac{16 !}{8 ! ( 16 - 8 ) !}

4 x + 18 = 7 (x + 3)

What are the solutions to the equation 18x^2+5=4 ?
The solutions to the equation 18x^2+5=4 are x= 1/(3sqrt(2))i,x=-1/(3sqrt(2))i
Find the zeros of 18x^2+5=4
The zeros of 18x^2+5=4 are x= 1/(3sqrt(2))i,x=-1/(3sqrt(2))i