What are the solutions to the equation 7z^2-28z=0 ?

The solutions to the equation 7z^2-28z=0 are z=4,z=0

Find the zeros of 7z^2-28z=0

The zeros of 7z^2-28z=0 are z=4,z=0

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7 z^{2} - 28 z = 0

z = 4, z = 0

Solution steps

7 z^{2} - 28 z = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 28 ) \pm ( - 28 ) ^{2} - 4 \cdot 7 \cdot 0}{2 \cdot 7}

(- 28)^{2} - 4 \cdot 7 \cdot 0 = 28

z_{1, 2} = \frac{- ( - 28 ) \pm 28}{2 \cdot 7}

Separate the solutions

z_{1} = \frac{- ( - 28 ) + 28}{2 \cdot 7}, z_{2} = \frac{- ( - 28 ) - 28}{2 \cdot 7}

z = \frac{- ( - 28 ) + 28}{2 \cdot 7} : 4

z = \frac{- ( - 28 ) - 28}{2 \cdot 7} : 0

The solutions to the quadratic equation are:

z = 4, z = 0

8 x + 8 y

\frac{3 w y ^{- 2}}{( w ^{- 1} y ) ^{3}}

t^{2} = 16

x^{2} + 11 x + 18 < 0

2 x^{2} - 5 x \leq 3

What are the solutions to the equation 7z^2-28z=0 ?
The solutions to the equation 7z^2-28z=0 are z=4,z=0
Find the zeros of 7z^2-28z=0
The zeros of 7z^2-28z=0 are z=4,z=0