What are the solutions to the equation x^2+13=-7x ?

The solutions to the equation x^2+13=-7x are x=-7/2+i(sqrt(3))/2 ,x=-7/2-i(sqrt(3))/2

Find the zeros of x^2+13=-7x

The zeros of x^2+13=-7x are x=-7/2+i(sqrt(3))/2 ,x=-7/2-i(sqrt(3))/2

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x^{2} + 13 = - 7 x

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

Solution steps

x^{2} + 13 = - 7 x

Move

7 x

to the left side

x^{2} + 13 + 7 x = 0

Write in the standard form

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

x^{2} + 7 x + 13 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 13}{2 \cdot 1}

Simplify

7^{2} - 4 \cdot 1 \cdot 13 : 3 i

x_{1, 2} = \frac{- 7 \pm 3 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 7 + 3 i}{2 \cdot 1}, x_{2} = \frac{- 7 - 3 i}{2 \cdot 1}

x = \frac{- 7 + 3 i}{2 \cdot 1} : - \frac{7}{2} + i \frac{3}{2}

x = \frac{- 7 - 3 i}{2 \cdot 1} : - \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}

lo g_{2} (2)

\frac{2}{5} (a + 3) - a \geq \frac{9}{5} (a - 2) - 3

\frac{2 x - 1}{5} - 5 = \frac{x + 4}{3} - \frac{7}{2}

7.13888 + 0.015528 x^{2} = 5

6 w - 90 = 48

What are the solutions to the equation x^2+13=-7x ?
The solutions to the equation x^2+13=-7x are x=-7/2+i(sqrt(3))/2 ,x=-7/2-i(sqrt(3))/2
Find the zeros of x^2+13=-7x
The zeros of x^2+13=-7x are x=-7/2+i(sqrt(3))/2 ,x=-7/2-i(sqrt(3))/2