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

The solutions to the equation x^2+10x+42=-3x+2 are x=-5,x=-8

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

The zeros of x^2+10x+42=-3x+2 are x=-5,x=-8

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

x = - 5, x = - 8

Solution steps

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

Move

2

to the left side

x^{2} + 10 x + 40 = - 3 x

Move

3 x

to the left side

x^{2} + 13 x + 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 1 \cdot 40}{2 \cdot 1}

1 3^{2} - 4 \cdot 1 \cdot 40 = 3

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

Separate the solutions

x_{1} = \frac{- 13 + 3}{2 \cdot 1}, x_{2} = \frac{- 13 - 3}{2 \cdot 1}

x = \frac{- 13 + 3}{2 \cdot 1} : - 5

x = \frac{- 13 - 3}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

x = - 5, x = - 8

\frac{- 5 - 2 i}{- 3 + 5 i}

1 = 7 x

\frac{π}{2} - \frac{4 π}{11}

z^{2} + 2 z - 8

What are the solutions to the equation x^2+10x+42=-3x+2 ?
The solutions to the equation x^2+10x+42=-3x+2 are x=-5,x=-8
Find the zeros of x^2+10x+42=-3x+2
The zeros of x^2+10x+42=-3x+2 are x=-5,x=-8