What are the solutions to the equation x^2+3x+47=-10x+7 ?

The solutions to the equation x^2+3x+47=-10x+7 are x=-5,x=-8

Find the zeros of x^2+3x+47=-10x+7

The zeros of x^2+3x+47=-10x+7 are x=-5,x=-8

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x^{2} + 3 x + 47 = - 10 x + 7

x = - 5, x = - 8

Solution steps

x^{2} + 3 x + 47 = - 10 x + 7

Move

7

to the left side

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

Move

10 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

5 x^{2} + 5 x + 5 = 0

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

\frac{2}{3} \cdot 24

4 x - x^{2}

\frac{1}{6} + \frac{3}{2}

What are the solutions to the equation x^2+3x+47=-10x+7 ?
The solutions to the equation x^2+3x+47=-10x+7 are x=-5,x=-8
Find the zeros of x^2+3x+47=-10x+7
The zeros of x^2+3x+47=-10x+7 are x=-5,x=-8