What are the solutions to the equation x^2-7x+13=0 ?

The solutions to the equation x^2-7x+13=0 are x= 7/2+i(sqrt(3))/2 ,x= 7/2-i(sqrt(3))/2

Find the zeros of x^2-7x+13=0

The zeros of x^2-7x+13=0 are x= 7/2+i(sqrt(3))/2 ,x= 7/2-i(sqrt(3))/2

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x^{2} - 7 x + 13 = 0

x = \frac{7}{2} + i \frac{3}{2}, x = \frac{7}{2} - i \frac{3}{2}

Solution steps

x^{2} - 7 x + 13 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 13}{2 \cdot 1}

Simplify

(- 7)^{2} - 4 \cdot 1 \cdot 13 : 3 i

x_{1, 2} = \frac{- ( - 7 ) \pm 3 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 3 i}{2 \cdot 1}, x_{2} = \frac{- ( - 7 ) - 3 i}{2 \cdot 1}

x = \frac{- ( - 7 ) + 3 i}{2 \cdot 1} : \frac{7}{2} + i \frac{3}{2}

x = \frac{- ( - 7 ) - 3 i}{2 \cdot 1} : \frac{7}{2} - i \frac{3}{2}

The solutions to the quadratic equation are:

x = \frac{7}{2} + i \frac{3}{2}, x = \frac{7}{2} - i \frac{3}{2}

\frac{x}{2} + 8 > 6

- 32 x + 48 = 0

\frac{1}{6} + \frac{1}{6} + \frac{1}{6}

(3 + 2 i) (5 - i) = 15 - 3 i + 10 i - 2 i^{2}

x^{2} - 5 x - x = 0

What are the solutions to the equation x^2-7x+13=0 ?
The solutions to the equation x^2-7x+13=0 are x= 7/2+i(sqrt(3))/2 ,x= 7/2-i(sqrt(3))/2
Find the zeros of x^2-7x+13=0
The zeros of x^2-7x+13=0 are x= 7/2+i(sqrt(3))/2 ,x= 7/2-i(sqrt(3))/2