What are the solutions to the equation 72x^2+96x=0 ?

The solutions to the equation 72x^2+96x=0 are x=0,x=-4/3

Find the zeros of 72x^2+96x=0

The zeros of 72x^2+96x=0 are x=0,x=-4/3

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72 x^{2} + 96 x = 0

x = 0, x = - \frac{4}{3}

Decimal Notation

x = 0, x = - 1.33333 \dots

Solution steps

72 x^{2} + 96 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 96 \pm 9 6 ^{2} - 4 \cdot 72 \cdot 0}{2 \cdot 72}

9 6^{2} - 4 \cdot 72 \cdot 0 = 96

x_{1, 2} = \frac{- 96 \pm 96}{2 \cdot 72}

Separate the solutions

x_{1} = \frac{- 96 + 96}{2 \cdot 72}, x_{2} = \frac{- 96 - 96}{2 \cdot 72}

x = \frac{- 96 + 96}{2 \cdot 72} : 0

x = \frac{- 96 - 96}{2 \cdot 72} : - \frac{4}{3}

The solutions to the quadratic equation are:

x = 0, x = - \frac{4}{3}

x^{2} + 6 x + 53 = 0

3 x^{2} + 10 x - 8 = 0

y^{2} + 4 y - 6 = 0

- 2 x^{2} - 20 x - 50 = 0

4.905 t^{2} - 90 t - 150 = 0

What are the solutions to the equation 72x^2+96x=0 ?
The solutions to the equation 72x^2+96x=0 are x=0,x=-4/3
Find the zeros of 72x^2+96x=0
The zeros of 72x^2+96x=0 are x=0,x=-4/3