What are the solutions to the equation z^2+24=11z ?

The solutions to the equation z^2+24=11z are z=8,z=3

Find the zeros of z^2+24=11z

The zeros of z^2+24=11z are z=8,z=3

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z^{2} + 24 = 11 z

z = 8, z = 3

Solution steps

z^{2} + 24 = 11 z

Move

11 z

to the left side

z^{2} + 24 - 11 z = 0

Write in the standard form

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

z^{2} - 11 z + 24 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 1 \cdot 24}{2 \cdot 1}

(- 11)^{2} - 4 \cdot 1 \cdot 24 = 5

z_{1, 2} = \frac{- ( - 11 ) \pm 5}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- ( - 11 ) + 5}{2 \cdot 1}, z_{2} = \frac{- ( - 11 ) - 5}{2 \cdot 1}

z = \frac{- ( - 11 ) + 5}{2 \cdot 1} : 8

z = \frac{- ( - 11 ) - 5}{2 \cdot 1} : 3

The solutions to the quadratic equation are:

z = 8, z = 3

t, s = - 15.45 t^{2} - 1.302 t + 5.234

3 x^{2} - 7 x = x^{2} - 14 x - 4

25 x^{2} + 36 = 60 x

m^{2} + 4 m + 1 = 0

- x^{2} + 3 x + 2

What are the solutions to the equation z^2+24=11z ?
The solutions to the equation z^2+24=11z are z=8,z=3
Find the zeros of z^2+24=11z
The zeros of z^2+24=11z are z=8,z=3