What are the solutions to the equation 5a^2-7=4 ?

The solutions to the equation 5a^2-7=4 are a=sqrt(11/5),a=-sqrt(11/5)

Find the zeros of 5a^2-7=4

The zeros of 5a^2-7=4 are a=sqrt(11/5),a=-sqrt(11/5)

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5 a^{2} - 7 = 4

a = \frac{11}{5}, a = - \frac{11}{5}

Decimal Notation

a = 1.48323 \dots, a = - 1.48323 \dots

Solution steps

5 a^{2} - 7 = 4

Move

7

to the right side

5 a^{2} = 11

Divide both sides by

5

a^{2} = \frac{11}{5}

For

x^{2} = f (a)

the solutions are

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

a = \frac{11}{5}, a = - \frac{11}{5}

- 3 + 4 x = 25

10 < r

2 (x - 3)^{2} + 6 = 14

\frac{\frac{4}{3}}{4}

2 x = 5 - 4 x^{2}

What are the solutions to the equation 5a^2-7=4 ?
The solutions to the equation 5a^2-7=4 are a=sqrt(11/5),a=-sqrt(11/5)
Find the zeros of 5a^2-7=4
The zeros of 5a^2-7=4 are a=sqrt(11/5),a=-sqrt(11/5)