What are the solutions to the equation 4x^2+89=-36x ?

The solutions to the equation 4x^2+89=-36x are x=-9/2+sqrt(2)i,x=-9/2-sqrt(2)i

Find the zeros of 4x^2+89=-36x

The zeros of 4x^2+89=-36x are x=-9/2+sqrt(2)i,x=-9/2-sqrt(2)i

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4 x^{2} + 89 = - 36 x

x = - \frac{9}{2} + 2 i, x = - \frac{9}{2} - 2 i

Solution steps

4 x^{2} + 89 = - 36 x

Move

36 x

to the left side

4 x^{2} + 89 + 36 x = 0

Write in the standard form

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

4 x^{2} + 36 x + 89 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 36 \pm 3 6 ^{2} - 4 \cdot 4 \cdot 89}{2 \cdot 4}

Simplify

3 6^{2} - 4 \cdot 4 \cdot 89 : 82 i

x_{1, 2} = \frac{- 36 \pm 8 2 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 36 + 8 2 i}{2 \cdot 4}, x_{2} = \frac{- 36 - 8 2 i}{2 \cdot 4}

x = \frac{- 36 + 8 2 i}{2 \cdot 4} : - \frac{9}{2} + 2 i

x = \frac{- 36 - 8 2 i}{2 \cdot 4} : - \frac{9}{2} - 2 i

The solutions to the quadratic equation are:

x = - \frac{9}{2} + 2 i, x = - \frac{9}{2} - 2 i

x^{2} - 2 x + 5

16 b^{2} - 1 = 0

x^{2} - 12 x + 1

5 (x^{2} - 4) + 10 = 10

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

What are the solutions to the equation 4x^2+89=-36x ?
The solutions to the equation 4x^2+89=-36x are x=-9/2+sqrt(2)i,x=-9/2-sqrt(2)i
Find the zeros of 4x^2+89=-36x
The zeros of 4x^2+89=-36x are x=-9/2+sqrt(2)i,x=-9/2-sqrt(2)i