What are the solutions to the equation 2x^2+7x=0 ?

The solutions to the equation 2x^2+7x=0 are x=0,x=-7/2

Find the zeros of 2x^2+7x=0

The zeros of 2x^2+7x=0 are x=0,x=-7/2

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2 x^{2} + 7 x = 0

x = 0, x = - \frac{7}{2}

Decimal Notation

x = 0, x = - 3.5

Solution steps

2 x^{2} + 7 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

7^{2} - 4 \cdot 2 \cdot 0 = 7

x_{1, 2} = \frac{- 7 \pm 7}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 7 + 7}{2 \cdot 2}, x_{2} = \frac{- 7 - 7}{2 \cdot 2}

x = \frac{- 7 + 7}{2 \cdot 2} : 0

x = \frac{- 7 - 7}{2 \cdot 2} : - \frac{7}{2}

The solutions to the quadratic equation are:

x = 0, x = - \frac{7}{2}

32 x - 50 x^{11}

\frac{x - 1}{x + 6} \geq \frac{x - 2}{x + 4}

3 x^{4} - 3

5 x - 4 \leq - 3 or 4 x + 6 > 0

(x + 10) (x - 2) = 0

What are the solutions to the equation 2x^2+7x=0 ?
The solutions to the equation 2x^2+7x=0 are x=0,x=-7/2
Find the zeros of 2x^2+7x=0
The zeros of 2x^2+7x=0 are x=0,x=-7/2