What are the solutions to the equation x^2+19x=0 ?

The solutions to the equation x^2+19x=0 are x=0,x=-19

Find the zeros of x^2+19x=0

The zeros of x^2+19x=0 are x=0,x=-19

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x^{2} + 19 x = 0

x = 0, x = - 19

Solution steps

x^{2} + 19 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 19 \pm 1 9 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

1 9^{2} - 4 \cdot 1 \cdot 0 = 19

x_{1, 2} = \frac{- 19 \pm 19}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 19 + 19}{2 \cdot 1}, x_{2} = \frac{- 19 - 19}{2 \cdot 1}

x = \frac{- 19 + 19}{2 \cdot 1} : 0

x = \frac{- 19 - 19}{2 \cdot 1} : - 19

The solutions to the quadratic equation are:

x = 0, x = - 19

- 6 + \frac{1}{4}

- 3 x = - 6

(t - 3)^{\frac{7}{2}} - (t - 3)^{\frac{3}{2}}

(p + 1)^{2} = 11 (p + 1) - 30

2^{n} + 2^{n}

What are the solutions to the equation x^2+19x=0 ?
The solutions to the equation x^2+19x=0 are x=0,x=-19
Find the zeros of x^2+19x=0
The zeros of x^2+19x=0 are x=0,x=-19