What are the solutions to the equation 9r^2+4=-109 ?

The solutions to the equation 9r^2+4=-109 are r=(sqrt(113))/3 i,r=-(sqrt(113))/3 i

Find the zeros of 9r^2+4=-109

The zeros of 9r^2+4=-109 are r=(sqrt(113))/3 i,r=-(sqrt(113))/3 i

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9 r^{2} + 4 = - 109

r = \frac{113}{3} i, r = - \frac{113}{3} i

Solution steps

9 r^{2} + 4 = - 109

Move

4

to the right side

9 r^{2} = - 113

Divide both sides by

9

r^{2} = - \frac{113}{9}

For

x^{2} = f (a)

the solutions are

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

r = - \frac{113}{9}, r = - - \frac{113}{9}

- \frac{113}{9} = \frac{113}{3} i

- - \frac{113}{9} = - \frac{113}{3} i

r = \frac{113}{3} i, r = - \frac{113}{3} i

\frac{8 ^{7}}{8 ^{7}}

6^{2} - 5 (6) - 2

10 m = 5 m - 20

20!

y^{2} - 24 y + 144 = - 100

What are the solutions to the equation 9r^2+4=-109 ?
The solutions to the equation 9r^2+4=-109 are r=(sqrt(113))/3 i,r=-(sqrt(113))/3 i
Find the zeros of 9r^2+4=-109
The zeros of 9r^2+4=-109 are r=(sqrt(113))/3 i,r=-(sqrt(113))/3 i