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

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

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

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

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0 = - x^{2} + 12 x

x = 0, x = 12

Solution steps

0 = - x^{2} + 12 x

Switch sides

- 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

x^{2} = 6 x - 117

y, x^{2} + y^{2} + 18 y + 81 = 18 x + 2 x y

- 4 x^{2} - 7 x + 36 = 0

4 x^{2} + 23 x = - 28

2 x^{2} + 3 x + 0.5 = 0

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