What are the solutions to the equation 25(x^2-3)+4=-7 ?

The solutions to the equation 25(x^2-3)+4=-7 are x= 8/5 ,x=-8/5

Find the zeros of 25(x^2-3)+4=-7

The zeros of 25(x^2-3)+4=-7 are x= 8/5 ,x=-8/5

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25 (x^{2} - 3) + 4 = - 7

x = \frac{8}{5}, x = - \frac{8}{5}

Decimal Notation

x = 1.6, x = - 1.6

Solution steps

25 (x^{2} - 3) + 4 = - 7

Move

4

to the right side

25 (x^{2} - 3) = - 11

Divide both sides by

25

x^{2} - 3 = - \frac{11}{25}

Move

3

to the right side

x^{2} = \frac{64}{25}

For

x^{2} = f (a)

the solutions are

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

x = \frac{64}{25}, x = - \frac{64}{25}

\frac{64}{25} = \frac{8}{5}

- \frac{64}{25} = - \frac{8}{5}

x = \frac{8}{5}, x = - \frac{8}{5}

4 r^{2} - 4 r + 1 = 0

(3 x - 1)^{2} = 64

x^{2} - 8 x + 16 = 0

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

3 x^{2} + 18 x + 5

What are the solutions to the equation 25(x^2-3)+4=-7 ?
The solutions to the equation 25(x^2-3)+4=-7 are x= 8/5 ,x=-8/5
Find the zeros of 25(x^2-3)+4=-7
The zeros of 25(x^2-3)+4=-7 are x= 8/5 ,x=-8/5