What are the solutions to the equation x^2+4x-13=0 ?

The solutions to the equation x^2+4x-13=0 are x=-2+sqrt(17),x=-2-sqrt(17)

Find the zeros of x^2+4x-13=0

The zeros of x^2+4x-13=0 are x=-2+sqrt(17),x=-2-sqrt(17)

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x^{2} + 4 x - 13 = 0

x = - 2 + 17, x = - 2 - 17

Solution steps

x^{2} + 4 x - 13 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot (- 13) = 217

x_{1, 2} = \frac{- 4 \pm 2 17}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 2 17}{2 \cdot 1}, x_{2} = \frac{- 4 - 2 17}{2 \cdot 1}

x = \frac{- 4 + 2 17}{2 \cdot 1} : - 2 + 17

x = \frac{- 4 - 2 17}{2 \cdot 1} : - 2 - 17

The solutions to the quadratic equation are:

x = - 2 + 17, x = - 2 - 17

x^{2} + 12 x = - 20

10 p^{2} - 2 p + 12 = 2

x^{2} - 5 x = 3

x^{2} - 12 x + 35 = 1760

12 x^{2} + 4 = 0

What are the solutions to the equation x^2+4x-13=0 ?
The solutions to the equation x^2+4x-13=0 are x=-2+sqrt(17),x=-2-sqrt(17)
Find the zeros of x^2+4x-13=0
The zeros of x^2+4x-13=0 are x=-2+sqrt(17),x=-2-sqrt(17)