What are the solutions to the equation z^2+40=14z ?

The solutions to the equation z^2+40=14z are z=10,z=4

Find the zeros of z^2+40=14z

The zeros of z^2+40=14z are z=10,z=4

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z^{2} + 40 = 14 z

z = 10, z = 4

Solution steps

z^{2} + 40 = 14 z

Move

14 z

to the left side

z^{2} + 40 - 14 z = 0

Write in the standard form

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

z^{2} - 14 z + 40 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 1 \cdot 40}{2 \cdot 1}

(- 14)^{2} - 4 \cdot 1 \cdot 40 = 6

z_{1, 2} = \frac{- ( - 14 ) \pm 6}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- ( - 14 ) + 6}{2 \cdot 1}, z_{2} = \frac{- ( - 14 ) - 6}{2 \cdot 1}

z = \frac{- ( - 14 ) + 6}{2 \cdot 1} : 10

z = \frac{- ( - 14 ) - 6}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

z = 10, z = 4

\frac{3}{9} x - \frac{1}{8} < - 5

- (2)^{4}

18 x y - 4 x - 9 y + 2

8 a c - 2 b d + b c - 16 a d

\frac{4 a b ^{5}}{a ^{3} b} \cdot \frac{3 a ^{5} b ^{2}}{8 a}

What are the solutions to the equation z^2+40=14z ?
The solutions to the equation z^2+40=14z are z=10,z=4
Find the zeros of z^2+40=14z
The zeros of z^2+40=14z are z=10,z=4