What are the solutions to the equation 25r^2+3=4 ?

The solutions to the equation 25r^2+3=4 are r= 1/5 ,r=-1/5

Find the zeros of 25r^2+3=4

The zeros of 25r^2+3=4 are r= 1/5 ,r=-1/5

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25 r^{2} + 3 = 4

r = \frac{1}{5}, r = - \frac{1}{5}

Solution steps

25 r^{2} + 3 = 4

Move

3

to the right side

25 r^{2} = 1

Divide both sides by

25

r^{2} = \frac{1}{25}

For

x^{2} = f (a)

the solutions are

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

r = \frac{1}{25}, r = - \frac{1}{25}

\frac{1}{25} = \frac{1}{5}

- \frac{1}{25} = - \frac{1}{5}

r = \frac{1}{5}, r = - \frac{1}{5}

4 x^{2} - 169 = 0

a^{2} + 2 a - 5 = 0

3 x = \frac{1}{5} x^{2} + 10

x^{2} - 18 x - 360 = 0

w, x yz + x z w - yz w + w^{2} = 5

What are the solutions to the equation 25r^2+3=4 ?
The solutions to the equation 25r^2+3=4 are r= 1/5 ,r=-1/5
Find the zeros of 25r^2+3=4
The zeros of 25r^2+3=4 are r= 1/5 ,r=-1/5