What are the solutions to the equation x^2+7/4 =-x ?

The solutions to the equation x^2+7/4 =-x are x=-1/2+isqrt(3/2),x=-1/2-isqrt(3/2)

Find the zeros of x^2+7/4 =-x

The zeros of x^2+7/4 =-x are x=-1/2+isqrt(3/2),x=-1/2-isqrt(3/2)

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x^{2} + \frac{7}{4} = - x

x = - \frac{1}{2} + i \frac{3}{2}, x = - \frac{1}{2} - i \frac{3}{2}

Solution steps

x^{2} + \frac{7}{4} = - x

Multiply both sides by

4

4 x^{2} + 7 = - 4 x

Move

4 x

to the left side

4 x^{2} + 7 + 4 x = 0

Write in the standard form

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

4 x^{2} + 4 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 4 \cdot 7}{2 \cdot 4}

Simplify

4^{2} - 4 \cdot 4 \cdot 7 : 46 i

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

Separate the solutions

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

x = \frac{- 4 + 4 6 i}{2 \cdot 4} : - \frac{1}{2} + i \frac{3}{2}

x = \frac{- 4 - 4 6 i}{2 \cdot 4} : - \frac{1}{2} - i \frac{3}{2}

The solutions to the quadratic equation are:

x = - \frac{1}{2} + i \frac{3}{2}, x = - \frac{1}{2} - i \frac{3}{2}

12 > - 2 x + 2 > - 6

5 x^{2} - 7 x + 3 = 0

\frac{π}{12} + 2 π

(3^{11}) (4^{3}) (4^{- 9}) (3^{5})

\frac{3 ( 8 - x )}{5} = x

What are the solutions to the equation x^2+7/4 =-x ?
The solutions to the equation x^2+7/4 =-x are x=-1/2+isqrt(3/2),x=-1/2-isqrt(3/2)
Find the zeros of x^2+7/4 =-x
The zeros of x^2+7/4 =-x are x=-1/2+isqrt(3/2),x=-1/2-isqrt(3/2)