What are the solutions to the equation roots x^2+2x+10 ?

The solutions to the equation roots x^2+2x+10 are x=-1+3i,x=-1-3i

Find the zeros of roots x^2+2x+10

The zeros of roots x^2+2x+10 are x=-1+3i,x=-1-3i

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roots

x^{2} + 2 x + 10

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

Solution steps

x^{2} + 2 x + 10

The roots are the intercepts with the x-axis

(y = 0)

x^{2} + 2 x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 10}{2 \cdot 1}

Simplify

2^{2} - 4 \cdot 1 \cdot 10 : 6 i

x_{1, 2} = \frac{- 2 \pm 6 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 6 i}{2 \cdot 1}, x_{2} = \frac{- 2 - 6 i}{2 \cdot 1}

x = \frac{- 2 + 6 i}{2 \cdot 1} : - 1 + 3 i

x = \frac{- 2 - 6 i}{2 \cdot 1} : - 1 - 3 i

The solutions to the quadratic equation are:

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

14 + 3 x^{2} = 26

x, 3 x^{2} + 12 x = 0

x^{2}

6 x^{2} + 8 x + 9

x, f = a (x - h)^{2} + k

What are the solutions to the equation roots x^2+2x+10 ?
The solutions to the equation roots x^2+2x+10 are x=-1+3i,x=-1-3i
Find the zeros of roots x^2+2x+10
The zeros of roots x^2+2x+10 are x=-1+3i,x=-1-3i