What are the solutions to the equation 4x^2+57=-28x ?

The solutions to the equation 4x^2+57=-28x are x=-7/2+sqrt(2)i,x=-7/2-sqrt(2)i

Find the zeros of 4x^2+57=-28x

The zeros of 4x^2+57=-28x are x=-7/2+sqrt(2)i,x=-7/2-sqrt(2)i

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4 x^{2} + 57 = - 28 x

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

Solution steps

4 x^{2} + 57 = - 28 x

Move

28 x

to the left side

4 x^{2} + 57 + 28 x = 0

Write in the standard form

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

4 x^{2} + 28 x + 57 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 4 \cdot 57}{2 \cdot 4}

Simplify

2 8^{2} - 4 \cdot 4 \cdot 57 : 82 i

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

Separate the solutions

x_{1} = \frac{- 28 + 8 2 i}{2 \cdot 4}, x_{2} = \frac{- 28 - 8 2 i}{2 \cdot 4}

x = \frac{- 28 + 8 2 i}{2 \cdot 4} : - \frac{7}{2} + 2 i

x = \frac{- 28 - 8 2 i}{2 \cdot 4} : - \frac{7}{2} - 2 i

The solutions to the quadratic equation are:

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

3 x^{2} - 136 x + 1300 = 0

4 x^{2} - 6 x + 2 = 5 + 5 x

- 43 = 5 - 3 b^{2}

- 4.9 t^{2} + 12.25 t + 115

x^{2} + 8 = - 17

What are the solutions to the equation 4x^2+57=-28x ?
The solutions to the equation 4x^2+57=-28x are x=-7/2+sqrt(2)i,x=-7/2-sqrt(2)i
Find the zeros of 4x^2+57=-28x
The zeros of 4x^2+57=-28x are x=-7/2+sqrt(2)i,x=-7/2-sqrt(2)i