What are the solutions to the equation 3x^2+x+1=-5x ?

The solutions to the equation 3x^2+x+1=-5x are x=(-3+sqrt(6))/3 ,x=-(3+sqrt(6))/3

Find the zeros of 3x^2+x+1=-5x

The zeros of 3x^2+x+1=-5x are x=(-3+sqrt(6))/3 ,x=-(3+sqrt(6))/3

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3 x^{2} + x + 1 = - 5 x

x = \frac{- 3 + 6}{3}, x = - \frac{3 + 6}{3}

Decimal Notation

x = - 0.18350 \dots, x = - 1.81649 \dots

Solution steps

3 x^{2} + x + 1 = - 5 x

Move

5 x

to the left side

3 x^{2} + 6 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 3 \cdot 1}{2 \cdot 3}

6^{2} - 4 \cdot 3 \cdot 1 = 26

x_{1, 2} = \frac{- 6 \pm 2 6}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 6 + 2 6}{2 \cdot 3}, x_{2} = \frac{- 6 - 2 6}{2 \cdot 3}

x = \frac{- 6 + 2 6}{2 \cdot 3} : \frac{- 3 + 6}{3}

x = \frac{- 6 - 2 6}{2 \cdot 3} : - \frac{3 + 6}{3}

The solutions to the quadratic equation are:

x = \frac{- 3 + 6}{3}, x = - \frac{3 + 6}{3}

(t + 2)^{2} + (t - 3)^{2} = 9

15 x^{2} + 21 x = 0

(x + 1) (x - 5) = 0

(x + 1)^{2} = 3

3 x^{2} - 5 = 13

What are the solutions to the equation 3x^2+x+1=-5x ?
The solutions to the equation 3x^2+x+1=-5x are x=(-3+sqrt(6))/3 ,x=-(3+sqrt(6))/3
Find the zeros of 3x^2+x+1=-5x
The zeros of 3x^2+x+1=-5x are x=(-3+sqrt(6))/3 ,x=-(3+sqrt(6))/3