What are the solutions to the equation (1/2)x^2+x+5=0 ?

The solutions to the equation (1/2)x^2+x+5=0 are x=-1+3i,x=-1-3i

Find the zeros of (1/2)x^2+x+5=0

The zeros of (1/2)x^2+x+5=0 are x=-1+3i,x=-1-3i

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

x = - 1 + 3 i, x = - 1 - 3 i

Solution steps

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

Refine

\frac{1}{2} x^{2} + x + 5 = 0

Write in the standard form

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

\frac{x ^{2}}{2} + x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot \frac{1}{2} \cdot 5}{2 \cdot \frac{1}{2}}

Simplify

1^{2} - 4 \cdot \frac{1}{2} \cdot 5 : 3 i

x_{1, 2} = \frac{- 1 \pm 3 i}{2 \cdot \frac{1}{2}}

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 1 + 3 i, x = - 1 - 3 i

\frac{2}{6} = \frac{u}{9}

\frac{2 ( x - 1 )}{3} - \frac{x + 4}{15} + 1 = x - \frac{3 ( x - 2 )}{5}

\frac{2 a - 3}{6} = \frac{2 a}{3} + \frac{1}{2}

y^{3} - x^{3} y^{4}

What are the solutions to the equation (1/2)x^2+x+5=0 ?
The solutions to the equation (1/2)x^2+x+5=0 are x=-1+3i,x=-1-3i
Find the zeros of (1/2)x^2+x+5=0
The zeros of (1/2)x^2+x+5=0 are x=-1+3i,x=-1-3i