What are the solutions to the equation 25r^2+7=0 ?

The solutions to the equation 25r^2+7=0 are r=(sqrt(7))/5 i,r=-(sqrt(7))/5 i

Find the zeros of 25r^2+7=0

The zeros of 25r^2+7=0 are r=(sqrt(7))/5 i,r=-(sqrt(7))/5 i

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25 r^{2} + 7 = 0

r = \frac{7}{5} i, r = - \frac{7}{5} i

Solution steps

25 r^{2} + 7 = 0

Move

7

to the right side

25 r^{2} = - 7

Divide both sides by

25

r^{2} = - \frac{7}{25}

For

x^{2} = f (a)

the solutions are

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

r = - \frac{7}{25}, r = - - \frac{7}{25}

- \frac{7}{25} = \frac{7}{5} i

- - \frac{7}{25} = - \frac{7}{5} i

r = \frac{7}{5} i, r = - \frac{7}{5} i

x^{2} + 12 x + 48 = - x + 6

2 (- 2) + 1

4 x^{2} - 4 x = - 7

∣ x - 4 ∣ = 0

7 (2 y - 8) = 14

What are the solutions to the equation 25r^2+7=0 ?
The solutions to the equation 25r^2+7=0 are r=(sqrt(7))/5 i,r=-(sqrt(7))/5 i
Find the zeros of 25r^2+7=0
The zeros of 25r^2+7=0 are r=(sqrt(7))/5 i,r=-(sqrt(7))/5 i