What are the solutions to the equation 4(x^2-7)+27=-12x^2 ?

The solutions to the equation 4(x^2-7)+27=-12x^2 are x= 1/4 ,x=-1/4

Find the zeros of 4(x^2-7)+27=-12x^2

The zeros of 4(x^2-7)+27=-12x^2 are x= 1/4 ,x=-1/4

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4 (x^{2} - 7) + 27 = - 12 x^{2}

x = \frac{1}{4}, x = - \frac{1}{4}

Solution steps

4 (x^{2} - 7) + 27 = - 12 x^{2}

Expand

4 (x^{2} - 7) + 27 : 4 x^{2} - 1

4 x^{2} - 1 = - 12 x^{2}

Move

1

to the right side

4 x^{2} = - 12 x^{2} + 1

Move

12 x^{2}

to the left side

16 x^{2} = 1

Divide both sides by

16

x^{2} = \frac{1}{16}

For

x^{2} = f (a)

the solutions are

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

x = \frac{1}{16}, x = - \frac{1}{16}

\frac{1}{16} = \frac{1}{4}

- \frac{1}{16} = - \frac{1}{4}

x = \frac{1}{4}, x = - \frac{1}{4}

x^{2} - 3 x = 6

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

x^{2} + 26 x

- 10 x^{2} + x + 11 = 0

8 x^{2} + 32 x = 0

What are the solutions to the equation 4(x^2-7)+27=-12x^2 ?
The solutions to the equation 4(x^2-7)+27=-12x^2 are x= 1/4 ,x=-1/4
Find the zeros of 4(x^2-7)+27=-12x^2
The zeros of 4(x^2-7)+27=-12x^2 are x= 1/4 ,x=-1/4