What are the solutions to the equation 4x^2+53=-28x ?

The solutions to the equation 4x^2+53=-28x are x=-7/2+i,x=-7/2-i

Find the zeros of 4x^2+53=-28x

The zeros of 4x^2+53=-28x are x=-7/2+i,x=-7/2-i

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4 x^{2} + 53 = - 28 x

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

Solution steps

4 x^{2} + 53 = - 28 x

Move

28 x

to the left side

4 x^{2} + 53 + 28 x = 0

Write in the standard form

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

4 x^{2} + 28 x + 53 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 4 \cdot 53}{2 \cdot 4}

Simplify

2 8^{2} - 4 \cdot 4 \cdot 53 : 8 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

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

x^{2} - 20 x = - 5 x - 13 - x^{2}

5 c^{2} - 21 c + 11 = - 4 c

100 = - 1500 + 12 x - 0.004 x^{2}

- 2 p^{2} + 13 p - 12 = - 6 p^{2}

What are the solutions to the equation 4x^2+53=-28x ?
The solutions to the equation 4x^2+53=-28x are x=-7/2+i,x=-7/2-i
Find the zeros of 4x^2+53=-28x
The zeros of 4x^2+53=-28x are x=-7/2+i,x=-7/2-i