What are the solutions to the equation x^2+20x+2600=0 ?

The solutions to the equation x^2+20x+2600=0 are x=-10+50i,x=-10-50i

Find the zeros of x^2+20x+2600=0

The zeros of x^2+20x+2600=0 are x=-10+50i,x=-10-50i

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x^{2} + 20 x + 2600 = 0

x = - 10 + 50 i, x = - 10 - 50 i

Solution steps

x^{2} + 20 x + 2600 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 1 \cdot 2600}{2 \cdot 1}

Simplify

2 0^{2} - 4 \cdot 1 \cdot 2600 : 100 i

x_{1, 2} = \frac{- 20 \pm 100 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 20 + 100 i}{2 \cdot 1}, x_{2} = \frac{- 20 - 100 i}{2 \cdot 1}

x = \frac{- 20 + 100 i}{2 \cdot 1} : - 10 + 50 i

x = \frac{- 20 - 100 i}{2 \cdot 1} : - 10 - 50 i

The solutions to the quadratic equation are:

x = - 10 + 50 i, x = - 10 - 50 i

9 - 8 x - x^{2}

(x + 5)^{2} = 2 x^{2} + 14 x + 4

5 x^{2} - 15 x + 11 = 0

7 x^{2} - 6 x = - 1

x, x^{2} - 4 x + y^{2} - 6 y + 22 = 0

What are the solutions to the equation x^2+20x+2600=0 ?
The solutions to the equation x^2+20x+2600=0 are x=-10+50i,x=-10-50i
Find the zeros of x^2+20x+2600=0
The zeros of x^2+20x+2600=0 are x=-10+50i,x=-10-50i