What are the solutions to the equation x^2+13x+42=0 ?

The solutions to the equation x^2+13x+42=0 are x=-6,x=-7

Find the zeros of x^2+13x+42=0

The zeros of x^2+13x+42=0 are x=-6,x=-7

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x^{2} + 13 x + 42 = 0

x = - 6, x = - 7

Solution steps

x^{2} + 13 x + 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 1 \cdot 42}{2 \cdot 1}

1 3^{2} - 4 \cdot 1 \cdot 42 = 1

x_{1, 2} = \frac{- 13 \pm 1}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 13 + 1}{2 \cdot 1}, x_{2} = \frac{- 13 - 1}{2 \cdot 1}

x = \frac{- 13 + 1}{2 \cdot 1} : - 6

x = \frac{- 13 - 1}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

x = - 6, x = - 7

1 + 3 x = x - 5

\frac{2 + 6}{2}

\frac{f}{10} - 1 > 1

(\frac{x ^{2}}{3} + y^{3})^{2}

3 s^{4} + 16 s

What are the solutions to the equation x^2+13x+42=0 ?
The solutions to the equation x^2+13x+42=0 are x=-6,x=-7
Find the zeros of x^2+13x+42=0
The zeros of x^2+13x+42=0 are x=-6,x=-7