What are the solutions to the equation k^2-7k+50=3 ?

The solutions to the equation k^2-7k+50=3 are k= 7/2+i(sqrt(139))/2 ,k= 7/2-i(sqrt(139))/2

Find the zeros of k^2-7k+50=3

The zeros of k^2-7k+50=3 are k= 7/2+i(sqrt(139))/2 ,k= 7/2-i(sqrt(139))/2

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k^{2} - 7 k + 50 = 3

k = \frac{7}{2} + i \frac{139}{2}, k = \frac{7}{2} - i \frac{139}{2}

Solution steps

k^{2} - 7 k + 50 = 3

Move

3

to the left side

k^{2} - 7 k + 47 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 47}{2 \cdot 1}

Simplify

(- 7)^{2} - 4 \cdot 1 \cdot 47 : 139 i

k_{1, 2} = \frac{- ( - 7 ) \pm 139 i}{2 \cdot 1}

Separate the solutions

k_{1} = \frac{- ( - 7 ) + 139 i}{2 \cdot 1}, k_{2} = \frac{- ( - 7 ) - 139 i}{2 \cdot 1}

k = \frac{- ( - 7 ) + 139 i}{2 \cdot 1} : \frac{7}{2} + i \frac{139}{2}

k = \frac{- ( - 7 ) - 139 i}{2 \cdot 1} : \frac{7}{2} - i \frac{139}{2}

The solutions to the quadratic equation are:

k = \frac{7}{2} + i \frac{139}{2}, k = \frac{7}{2} - i \frac{139}{2}

2 y^{2} - 50 = 0

(4 - k) x^{2} - 2 k x = - 9

22 - 6 x^{2} = 0

w^{2} = - 144

2 r^{2} - 5 r - 3 = 0

What are the solutions to the equation k^2-7k+50=3 ?
The solutions to the equation k^2-7k+50=3 are k= 7/2+i(sqrt(139))/2 ,k= 7/2-i(sqrt(139))/2
Find the zeros of k^2-7k+50=3
The zeros of k^2-7k+50=3 are k= 7/2+i(sqrt(139))/2 ,k= 7/2-i(sqrt(139))/2