What are the solutions to the equation m^2-m+6=0 ?

The solutions to the equation m^2-m+6=0 are m= 1/2+i(sqrt(23))/2 ,m= 1/2-i(sqrt(23))/2

Find the zeros of m^2-m+6=0

The zeros of m^2-m+6=0 are m= 1/2+i(sqrt(23))/2 ,m= 1/2-i(sqrt(23))/2

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m^{2} - m + 6 = 0

m = \frac{1}{2} + i \frac{23}{2}, m = \frac{1}{2} - i \frac{23}{2}

Solution steps

m^{2} - m + 6 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

Simplify

(- 1)^{2} - 4 \cdot 1 \cdot 6 : 23 i

m_{1, 2} = \frac{- ( - 1 ) \pm 23 i}{2 \cdot 1}

Separate the solutions

m_{1} = \frac{- ( - 1 ) + 23 i}{2 \cdot 1}, m_{2} = \frac{- ( - 1 ) - 23 i}{2 \cdot 1}

m = \frac{- ( - 1 ) + 23 i}{2 \cdot 1} : \frac{1}{2} + i \frac{23}{2}

m = \frac{- ( - 1 ) - 23 i}{2 \cdot 1} : \frac{1}{2} - i \frac{23}{2}

The solutions to the quadratic equation are:

m = \frac{1}{2} + i \frac{23}{2}, m = \frac{1}{2} - i \frac{23}{2}

5 c^{2} - 36 c = 32

0 = 40 + 5 t - 4.9 t^{2}

4 x \cdot (2 x + 4) = - 8

2 x^{2} - 7 x = 12

6 x^{2} + 8 = - 16

What are the solutions to the equation m^2-m+6=0 ?
The solutions to the equation m^2-m+6=0 are m= 1/2+i(sqrt(23))/2 ,m= 1/2-i(sqrt(23))/2
Find the zeros of m^2-m+6=0
The zeros of m^2-m+6=0 are m= 1/2+i(sqrt(23))/2 ,m= 1/2-i(sqrt(23))/2