What are the solutions to the equation -x^2+12x=0 ?

The solutions to the equation -x^2+12x=0 are x=0,x=12

Find the zeros of -x^2+12x=0

The zeros of -x^2+12x=0 are x=0,x=12

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- x^{2} + 12 x = 0

x = 0, x = 12

Solution steps

- x^{2} + 12 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

1 2^{2} - 4 (- 1) \cdot 0 = 12

x_{1, 2} = \frac{- 12 \pm 12}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 12 + 12}{2 ( - 1 )}, x_{2} = \frac{- 12 - 12}{2 ( - 1 )}

x = \frac{- 12 + 12}{2 ( - 1 )} : 0

x = \frac{- 12 - 12}{2 ( - 1 )} : 12

The solutions to the quadratic equation are:

x = 0, x = 12

r^{2} - 2 r - 8 = 0

x^{2} = 2 x - 26

2 x^{2} - 7 x = 0

\frac{1}{1} - \frac{1}{2}

y^{2} - 11 y - 12

What are the solutions to the equation -x^2+12x=0 ?
The solutions to the equation -x^2+12x=0 are x=0,x=12
Find the zeros of -x^2+12x=0
The zeros of -x^2+12x=0 are x=0,x=12