What are the solutions to the equation 25(x^2-1)+6=-10 ?

The solutions to the equation 25(x^2-1)+6=-10 are x= 3/5 ,x=-3/5

Find the zeros of 25(x^2-1)+6=-10

The zeros of 25(x^2-1)+6=-10 are x= 3/5 ,x=-3/5

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25 (x^{2} - 1) + 6 = - 10

x = \frac{3}{5}, x = - \frac{3}{5}

Decimal Notation

x = 0.6, x = - 0.6

Solution steps

25 (x^{2} - 1) + 6 = - 10

Move

6

to the right side

25 (x^{2} - 1) = - 16

Divide both sides by

25

x^{2} - 1 = - \frac{16}{25}

Move

1

to the right side

x^{2} = \frac{9}{25}

For

x^{2} = f (a)

the solutions are

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

x = \frac{9}{25}, x = - \frac{9}{25}

\frac{9}{25} = \frac{3}{5}

- \frac{9}{25} = - \frac{3}{5}

x = \frac{3}{5}, x = - \frac{3}{5}

r - 8 \leq 7

7 y = 3 y^{2} + 2

6^{2} + 7^{2}

(x - 5) (x + 6) = 0

24 a^{2} b m^{2} - 36 b^{2} m^{4}

What are the solutions to the equation 25(x^2-1)+6=-10 ?
The solutions to the equation 25(x^2-1)+6=-10 are x= 3/5 ,x=-3/5
Find the zeros of 25(x^2-1)+6=-10
The zeros of 25(x^2-1)+6=-10 are x= 3/5 ,x=-3/5