What are the solutions to the equation 9p^2+10p-3=-7 ?

The solutions to the equation 9p^2+10p-3=-7 are p=-5/9+i(sqrt(11))/9 ,p=-5/9-i(sqrt(11))/9

Find the zeros of 9p^2+10p-3=-7

The zeros of 9p^2+10p-3=-7 are p=-5/9+i(sqrt(11))/9 ,p=-5/9-i(sqrt(11))/9

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9 p^{2} + 10 p - 3 = - 7

p = - \frac{5}{9} + i \frac{11}{9}, p = - \frac{5}{9} - i \frac{11}{9}

Solution steps

9 p^{2} + 10 p - 3 = - 7

Move

7

to the left side

9 p^{2} + 10 p + 4 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 9 \cdot 4}{2 \cdot 9}

Simplify

1 0^{2} - 4 \cdot 9 \cdot 4 : 211 i

p_{1, 2} = \frac{- 10 \pm 2 11 i}{2 \cdot 9}

Separate the solutions

p_{1} = \frac{- 10 + 2 11 i}{2 \cdot 9}, p_{2} = \frac{- 10 - 2 11 i}{2 \cdot 9}

p = \frac{- 10 + 2 11 i}{2 \cdot 9} : - \frac{5}{9} + i \frac{11}{9}

p = \frac{- 10 - 2 11 i}{2 \cdot 9} : - \frac{5}{9} - i \frac{11}{9}

The solutions to the quadratic equation are:

p = - \frac{5}{9} + i \frac{11}{9}, p = - \frac{5}{9} - i \frac{11}{9}

\frac{x}{4} \geq 8

(x - 2)^{2} = 8

x^{2} - 16 x + 15 = 0

\frac{1}{\frac{1}{6} + \frac{1}{6}}

2 n - 3 > 5

What are the solutions to the equation 9p^2+10p-3=-7 ?
The solutions to the equation 9p^2+10p-3=-7 are p=-5/9+i(sqrt(11))/9 ,p=-5/9-i(sqrt(11))/9
Find the zeros of 9p^2+10p-3=-7
The zeros of 9p^2+10p-3=-7 are p=-5/9+i(sqrt(11))/9 ,p=-5/9-i(sqrt(11))/9