What are the solutions to the equation 2x^2+5=-5x ?

The solutions to the equation 2x^2+5=-5x are x=-5/4+i(sqrt(15))/4 ,x=-5/4-i(sqrt(15))/4

Find the zeros of 2x^2+5=-5x

The zeros of 2x^2+5=-5x are x=-5/4+i(sqrt(15))/4 ,x=-5/4-i(sqrt(15))/4

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

x = - \frac{5}{4} + i \frac{15}{4}, x = - \frac{5}{4} - i \frac{15}{4}

Solution steps

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

Move

5 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

5^{2} - 4 \cdot 2 \cdot 5 : 15 i

x_{1, 2} = \frac{- 5 \pm 15 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 5 + 15 i}{2 \cdot 2}, x_{2} = \frac{- 5 - 15 i}{2 \cdot 2}

x = \frac{- 5 + 15 i}{2 \cdot 2} : - \frac{5}{4} + i \frac{15}{4}

x = \frac{- 5 - 15 i}{2 \cdot 2} : - \frac{5}{4} - i \frac{15}{4}

The solutions to the quadratic equation are:

x = - \frac{5}{4} + i \frac{15}{4}, x = - \frac{5}{4} - i \frac{15}{4}

ln (3) - ln (2)

∣ x - 5 ∣ = 3

x^{2} + 16 x + 59 = 0

\frac{1}{3} z - \frac{1}{6} = - \frac{2}{3} z - \frac{1}{3}

\frac{1}{20} + \frac{1}{30}

What are the solutions to the equation 2x^2+5=-5x ?
The solutions to the equation 2x^2+5=-5x are x=-5/4+i(sqrt(15))/4 ,x=-5/4-i(sqrt(15))/4
Find the zeros of 2x^2+5=-5x
The zeros of 2x^2+5=-5x are x=-5/4+i(sqrt(15))/4 ,x=-5/4-i(sqrt(15))/4