What are the solutions to the equation z^2+28=11z ?

The solutions to the equation z^2+28=11z are z=7,z=4

Find the zeros of z^2+28=11z

The zeros of z^2+28=11z are z=7,z=4

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z^{2} + 28 = 11 z

z = 7, z = 4

Solution steps

z^{2} + 28 = 11 z

Move

11 z

to the left side

z^{2} + 28 - 11 z = 0

Write in the standard form

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

z^{2} - 11 z + 28 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 1 \cdot 28}{2 \cdot 1}

(- 11)^{2} - 4 \cdot 1 \cdot 28 = 3

z_{1, 2} = \frac{- ( - 11 ) \pm 3}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- ( - 11 ) + 3}{2 \cdot 1}, z_{2} = \frac{- ( - 11 ) - 3}{2 \cdot 1}

z = \frac{- ( - 11 ) + 3}{2 \cdot 1} : 7

z = \frac{- ( - 11 ) - 3}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

z = 7, z = 4

384 x^{4} y^{3}

∣ 2 x + 4 ∣ > 7

\frac{15 !}{3 ! ( 15 - 3 ) !}

5 x - 10 = 0

- 5 x > 7

What are the solutions to the equation z^2+28=11z ?
The solutions to the equation z^2+28=11z are z=7,z=4
Find the zeros of z^2+28=11z
The zeros of z^2+28=11z are z=7,z=4