What are the solutions to the equation 4x^2+55=-28x ?

The solutions to the equation 4x^2+55=-28x are x=-7/2+isqrt(3/2),x=-7/2-isqrt(3/2)

Find the zeros of 4x^2+55=-28x

The zeros of 4x^2+55=-28x are x=-7/2+isqrt(3/2),x=-7/2-isqrt(3/2)

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4 x^{2} + 55 = - 28 x

x = - \frac{7}{2} + i \frac{3}{2}, x = - \frac{7}{2} - i \frac{3}{2}

Solution steps

4 x^{2} + 55 = - 28 x

Move

28 x

to the left side

4 x^{2} + 55 + 28 x = 0

Write in the standard form

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

4 x^{2} + 28 x + 55 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 4 \cdot 55}{2 \cdot 4}

Simplify

2 8^{2} - 4 \cdot 4 \cdot 55 : 46 i

x_{1, 2} = \frac{- 28 \pm 4 6 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 28 + 4 6 i}{2 \cdot 4}, x_{2} = \frac{- 28 - 4 6 i}{2 \cdot 4}

x = \frac{- 28 + 4 6 i}{2 \cdot 4} : - \frac{7}{2} + i \frac{3}{2}

x = \frac{- 28 - 4 6 i}{2 \cdot 4} : - \frac{7}{2} - i \frac{3}{2}

The solutions to the quadratic equation are:

x = - \frac{7}{2} + i \frac{3}{2}, x = - \frac{7}{2} - i \frac{3}{2}

- \frac{5}{2} + 6

e^{- y} + e^{- 2 x - y}

6 x^{2} + 8 x - 8 \leq 0

\frac{d}{d x} (a x^{2} + b x + c)

3 x^{2} + 18 x + 32 = 5

What are the solutions to the equation 4x^2+55=-28x ?
The solutions to the equation 4x^2+55=-28x are x=-7/2+isqrt(3/2),x=-7/2-isqrt(3/2)
Find the zeros of 4x^2+55=-28x
The zeros of 4x^2+55=-28x are x=-7/2+isqrt(3/2),x=-7/2-isqrt(3/2)