What are the solutions to the equation x^2+5x+25=0 ?

The solutions to the equation x^2+5x+25=0 are x=-5/2+i(5sqrt(3))/2 ,x=-5/2-i(5sqrt(3))/2

Find the zeros of x^2+5x+25=0

The zeros of x^2+5x+25=0 are x=-5/2+i(5sqrt(3))/2 ,x=-5/2-i(5sqrt(3))/2

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x^{2} + 5 x + 25 = 0

x = - \frac{5}{2} + i \frac{5 3}{2}, x = - \frac{5}{2} - i \frac{5 3}{2}

Solution steps

x^{2} + 5 x + 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 25}{2 \cdot 1}

Simplify

5^{2} - 4 \cdot 1 \cdot 25 : 53 i

x_{1, 2} = \frac{- 5 \pm 5 3 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 5 + 5 3 i}{2 \cdot 1}, x_{2} = \frac{- 5 - 5 3 i}{2 \cdot 1}

x = \frac{- 5 + 5 3 i}{2 \cdot 1} : - \frac{5}{2} + i \frac{5 3}{2}

x = \frac{- 5 - 5 3 i}{2 \cdot 1} : - \frac{5}{2} - i \frac{5 3}{2}

The solutions to the quadratic equation are:

x = - \frac{5}{2} + i \frac{5 3}{2}, x = - \frac{5}{2} - i \frac{5 3}{2}

\frac{1}{6} (6)

1 + 3 ∣ 2 x - 5 ∣ = 4

x^{2} - 3 x y + \frac{1}{2} y^{2}

∣ 2 x + 5 ∣ < 13

\frac{( \frac{3}{4} ) + 3}{2} < \frac{2 ( \frac{3}{4} ) - 5}{4}

What are the solutions to the equation x^2+5x+25=0 ?
The solutions to the equation x^2+5x+25=0 are x=-5/2+i(5sqrt(3))/2 ,x=-5/2-i(5sqrt(3))/2
Find the zeros of x^2+5x+25=0
The zeros of x^2+5x+25=0 are x=-5/2+i(5sqrt(3))/2 ,x=-5/2-i(5sqrt(3))/2