What are the solutions to the equation x^2+10x+42=-3x ?

The solutions to the equation x^2+10x+42=-3x are x=-6,x=-7

Find the zeros of x^2+10x+42=-3x

The zeros of x^2+10x+42=-3x are x=-6,x=-7

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x^{2} + 10 x + 42 = - 3 x

x = - 6, x = - 7

Solution steps

x^{2} + 10 x + 42 = - 3 x

Move

3 x

to the left side

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

18 x^{2} y - 24 x^{2} - 12 y + 16

- 2 x^{2} + 4 x - 1 = 0

6 x - 2 (x + 9) = 2

(5^{3} x^{2} y^{4})^{0}

4 x + 5 x = 5 x + 7

What are the solutions to the equation x^2+10x+42=-3x ?
The solutions to the equation x^2+10x+42=-3x are x=-6,x=-7
Find the zeros of x^2+10x+42=-3x
The zeros of x^2+10x+42=-3x are x=-6,x=-7