What are the solutions to the equation z^2+36=13z ?

The solutions to the equation z^2+36=13z are z=9,z=4

Find the zeros of z^2+36=13z

The zeros of z^2+36=13z are z=9,z=4

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z^{2} + 36 = 13 z

z = 9, z = 4

Solution steps

z^{2} + 36 = 13 z

Move

13 z

to the left side

z^{2} + 36 - 13 z = 0

Write in the standard form

a x^{2} + b x + c = 0

z^{2} - 13 z + 36 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

(- 13)^{2} - 4 \cdot 1 \cdot 36 = 5

z_{1, 2} = \frac{- ( - 13 ) \pm 5}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- ( - 13 ) + 5}{2 \cdot 1}, z_{2} = \frac{- ( - 13 ) - 5}{2 \cdot 1}

z = \frac{- ( - 13 ) + 5}{2 \cdot 1} : 9

z = \frac{- ( - 13 ) - 5}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

z = 9, z = 4

- (6)^{- 1}

5 ((- 7) + 9) \leq 10

4 x + (- 6) = 10

5 i \cdot i

(x^{2} - 4) (x^{2} + 6 x + 9)

What are the solutions to the equation z^2+36=13z ?
The solutions to the equation z^2+36=13z are z=9,z=4
Find the zeros of z^2+36=13z
The zeros of z^2+36=13z are z=9,z=4