What are the solutions to the equation m^2-7m+32=8 ?

The solutions to the equation m^2-7m+32=8 are m= 7/2+i(sqrt(47))/2 ,m= 7/2-i(sqrt(47))/2

Find the zeros of m^2-7m+32=8

The zeros of m^2-7m+32=8 are m= 7/2+i(sqrt(47))/2 ,m= 7/2-i(sqrt(47))/2

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m^{2} - 7 m + 32 = 8

m = \frac{7}{2} + i \frac{47}{2}, m = \frac{7}{2} - i \frac{47}{2}

Solution steps

m^{2} - 7 m + 32 = 8

Move

8

to the left side

m^{2} - 7 m + 24 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 24}{2 \cdot 1}

Simplify

(- 7)^{2} - 4 \cdot 1 \cdot 24 : 47 i

m_{1, 2} = \frac{- ( - 7 ) \pm 47 i}{2 \cdot 1}

Separate the solutions

m_{1} = \frac{- ( - 7 ) + 47 i}{2 \cdot 1}, m_{2} = \frac{- ( - 7 ) - 47 i}{2 \cdot 1}

m = \frac{- ( - 7 ) + 47 i}{2 \cdot 1} : \frac{7}{2} + i \frac{47}{2}

m = \frac{- ( - 7 ) - 47 i}{2 \cdot 1} : \frac{7}{2} - i \frac{47}{2}

The solutions to the quadratic equation are:

m = \frac{7}{2} + i \frac{47}{2}, m = \frac{7}{2} - i \frac{47}{2}

27 x - 3 x^{3}

\frac{12 x ^{4} - 16 x ^{3} y ^{2}}{- 4 x ^{3}}

7 = \frac{1}{2} (x + 6)

170

a^{8}

What are the solutions to the equation m^2-7m+32=8 ?
The solutions to the equation m^2-7m+32=8 are m= 7/2+i(sqrt(47))/2 ,m= 7/2-i(sqrt(47))/2
Find the zeros of m^2-7m+32=8
The zeros of m^2-7m+32=8 are m= 7/2+i(sqrt(47))/2 ,m= 7/2-i(sqrt(47))/2