What are the solutions to the equation x^2+7x+29=0 ?

The solutions to the equation x^2+7x+29=0 are x=-7/2+i(sqrt(67))/2 ,x=-7/2-i(sqrt(67))/2

Find the zeros of x^2+7x+29=0

The zeros of x^2+7x+29=0 are x=-7/2+i(sqrt(67))/2 ,x=-7/2-i(sqrt(67))/2

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x^{2} + 7 x + 29 = 0

x = - \frac{7}{2} + i \frac{67}{2}, x = - \frac{7}{2} - i \frac{67}{2}

Solution steps

x^{2} + 7 x + 29 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 29}{2 \cdot 1}

Simplify

7^{2} - 4 \cdot 1 \cdot 29 : 67 i

x_{1, 2} = \frac{- 7 \pm 67 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 7 + 67 i}{2 \cdot 1}, x_{2} = \frac{- 7 - 67 i}{2 \cdot 1}

x = \frac{- 7 + 67 i}{2 \cdot 1} : - \frac{7}{2} + i \frac{67}{2}

x = \frac{- 7 - 67 i}{2 \cdot 1} : - \frac{7}{2} - i \frac{67}{2}

The solutions to the quadratic equation are:

x = - \frac{7}{2} + i \frac{67}{2}, x = - \frac{7}{2} - i \frac{67}{2}

x (4 x - 3) = 0

(x + \frac{5}{2})^{2} = \frac{5}{4}

x, x^{2} y^{2} + 3 y = 4 x

x^{2} + 17 = 13

x^{2} + 121 = 22 x

What are the solutions to the equation x^2+7x+29=0 ?
The solutions to the equation x^2+7x+29=0 are x=-7/2+i(sqrt(67))/2 ,x=-7/2-i(sqrt(67))/2
Find the zeros of x^2+7x+29=0
The zeros of x^2+7x+29=0 are x=-7/2+i(sqrt(67))/2 ,x=-7/2-i(sqrt(67))/2