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

The solutions to the equation -2x^2+48x=0 are x=0,x=24

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

The zeros of -2x^2+48x=0 are x=0,x=24

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

x = 0, x = 24

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

x = \frac{- 48 - 48}{2 ( - 2 )} : 24

The solutions to the quadratic equation are:

x = 0, x = 24

y, x = \frac{1}{4} y^{2} + \frac{1}{2}

a^{2} + 7^{2} = 8^{2}

x^{2} - 4 x + 9 = 0

x^{2} + 18 x + 18 = - 5

- 12 x^{2} - 30 x = 0

What are the solutions to the equation -2x^2+48x=0 ?
The solutions to the equation -2x^2+48x=0 are x=0,x=24
Find the zeros of -2x^2+48x=0
The zeros of -2x^2+48x=0 are x=0,x=24