What are the solutions to the equation -16x^2+48x=0 ?

The solutions to the equation -16x^2+48x=0 are x=0,x=3

Find the zeros of -16x^2+48x=0

The zeros of -16x^2+48x=0 are x=0,x=3

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- 16 x^{2} + 48 x = 0

x = 0, x = 3

Solution steps

- 16 x^{2} + 48 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 48 \pm 4 8 ^{2} - 4 ( - 16 ) \cdot 0}{2 ( - 16 )}

4 8^{2} - 4 (- 16) \cdot 0 = 48

x_{1, 2} = \frac{- 48 \pm 48}{2 ( - 16 )}

Separate the solutions

x_{1} = \frac{- 48 + 48}{2 ( - 16 )}, x_{2} = \frac{- 48 - 48}{2 ( - 16 )}

x = \frac{- 48 + 48}{2 ( - 16 )} : 0

x = \frac{- 48 - 48}{2 ( - 16 )} : 3

The solutions to the quadratic equation are:

x = 0, x = 3

\frac{18 !}{5 ! ( 18 - 5 ) !}

\frac{i + 2}{i - 2}

6 (\frac{1}{3} d + 4) > 2 (d + 12)

(2 w - 1) w = 21

(x - 2)^{3} + 8 y^{3}

What are the solutions to the equation -16x^2+48x=0 ?
The solutions to the equation -16x^2+48x=0 are x=0,x=3
Find the zeros of -16x^2+48x=0
The zeros of -16x^2+48x=0 are x=0,x=3