What are the solutions to the equation 2q^2+9=-q ?

The solutions to the equation 2q^2+9=-q are q=-1/4+i(sqrt(71))/4 ,q=-1/4-i(sqrt(71))/4

Find the zeros of 2q^2+9=-q

The zeros of 2q^2+9=-q are q=-1/4+i(sqrt(71))/4 ,q=-1/4-i(sqrt(71))/4

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2 q^{2} + 9 = - q

q = - \frac{1}{4} + i \frac{71}{4}, q = - \frac{1}{4} - i \frac{71}{4}

Solution steps

2 q^{2} + 9 = - q

Move

q

to the left side

2 q^{2} + 9 + q = 0

Write in the standard form

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

2 q^{2} + q + 9 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 \cdot 9}{2 \cdot 2}

Simplify

1^{2} - 4 \cdot 2 \cdot 9 : 71 i

q_{1, 2} = \frac{- 1 \pm 71 i}{2 \cdot 2}

Separate the solutions

q_{1} = \frac{- 1 + 71 i}{2 \cdot 2}, q_{2} = \frac{- 1 - 71 i}{2 \cdot 2}

q = \frac{- 1 + 71 i}{2 \cdot 2} : - \frac{1}{4} + i \frac{71}{4}

q = \frac{- 1 - 71 i}{2 \cdot 2} : - \frac{1}{4} - i \frac{71}{4}

The solutions to the quadratic equation are:

q = - \frac{1}{4} + i \frac{71}{4}, q = - \frac{1}{4} - i \frac{71}{4}

(x + 3) (x + 7) = 11

2 c^{2} - 3 c - 9 = 0

(3 x + 5)^{2} - \frac{1}{3} = 0

X^{2} - 4 X + 1 = 0

2 x^{2} + 12 x - 10 = 0

What are the solutions to the equation 2q^2+9=-q ?
The solutions to the equation 2q^2+9=-q are q=-1/4+i(sqrt(71))/4 ,q=-1/4-i(sqrt(71))/4
Find the zeros of 2q^2+9=-q
The zeros of 2q^2+9=-q are q=-1/4+i(sqrt(71))/4 ,q=-1/4-i(sqrt(71))/4