What are the solutions to the equation 7r^2-5=2r+9r^2 ?

The solutions to the equation 7r^2-5=2r+9r^2 are r=-1/2-i 3/2 ,r=-1/2+i 3/2

Find the zeros of 7r^2-5=2r+9r^2

The zeros of 7r^2-5=2r+9r^2 are r=-1/2-i 3/2 ,r=-1/2+i 3/2

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7 r^{2} - 5 = 2 r + 9 r^{2}

r = - \frac{1}{2} - i \frac{3}{2}, r = - \frac{1}{2} + i \frac{3}{2}

Solution steps

7 r^{2} - 5 = 2 r + 9 r^{2}

Move

9 r^{2}

to the left side

- 2 r^{2} - 5 = 2 r

Move

2 r

to the left side

- 2 r^{2} - 5 - 2 r = 0

Write in the standard form

a x^{2} + b x + c = 0

- 2 r^{2} - 2 r - 5 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 2 ) ( - 5 )}{2 ( - 2 )}

Simplify

(- 2)^{2} - 4 (- 2) (- 5) : 6 i

r_{1, 2} = \frac{- ( - 2 ) \pm 6 i}{2 ( - 2 )}

Separate the solutions

r_{1} = \frac{- ( - 2 ) + 6 i}{2 ( - 2 )}, r_{2} = \frac{- ( - 2 ) - 6 i}{2 ( - 2 )}

r = \frac{- ( - 2 ) + 6 i}{2 ( - 2 )} : - \frac{1}{2} - i \frac{3}{2}

r = \frac{- ( - 2 ) - 6 i}{2 ( - 2 )} : - \frac{1}{2} + i \frac{3}{2}

The solutions to the quadratic equation are:

r = - \frac{1}{2} - i \frac{3}{2}, r = - \frac{1}{2} + i \frac{3}{2}

(\frac{1}{h ^{- 2}})^{- 1} \cdot h^{3}

\frac{3}{2} + 8

\frac{3}{8} r - 6 < 15

1 0^{2} + 1 5^{2} = x^{2}

a^{2} + a - 20

What are the solutions to the equation 7r^2-5=2r+9r^2 ?
The solutions to the equation 7r^2-5=2r+9r^2 are r=-1/2-i 3/2 ,r=-1/2+i 3/2
Find the zeros of 7r^2-5=2r+9r^2
The zeros of 7r^2-5=2r+9r^2 are r=-1/2-i 3/2 ,r=-1/2+i 3/2