What are the solutions to the equation 3x^2+8x=-1 ?

The solutions to the equation 3x^2+8x=-1 are x=(-4+sqrt(13))/3 ,x=-(4+sqrt(13))/3

Find the zeros of 3x^2+8x=-1

The zeros of 3x^2+8x=-1 are x=(-4+sqrt(13))/3 ,x=-(4+sqrt(13))/3

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3 x^{2} + 8 x = - 1

x = \frac{- 4 + 13}{3}, x = - \frac{4 + 13}{3}

Decimal Notation

x = - 0.13148 \dots, x = - 2.53518 \dots

Solution steps

3 x^{2} + 8 x = - 1

Move

1

to the left side

3 x^{2} + 8 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 3 \cdot 1}{2 \cdot 3}

8^{2} - 4 \cdot 3 \cdot 1 = 213

x_{1, 2} = \frac{- 8 \pm 2 13}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 8 + 2 13}{2 \cdot 3}, x_{2} = \frac{- 8 - 2 13}{2 \cdot 3}

x = \frac{- 8 + 2 13}{2 \cdot 3} : \frac{- 4 + 13}{3}

x = \frac{- 8 - 2 13}{2 \cdot 3} : - \frac{4 + 13}{3}

The solutions to the quadratic equation are:

x = \frac{- 4 + 13}{3}, x = - \frac{4 + 13}{3}

11.3 > \frac{b}{4.3}

\frac{1}{2} - \frac{x}{4} = \frac{1}{8}

10 - x < 4 x \leq 25 - x

5 c + 20 c^{2} + 20 c^{3}

\frac{π}{2} + \frac{π}{4}

What are the solutions to the equation 3x^2+8x=-1 ?
The solutions to the equation 3x^2+8x=-1 are x=(-4+sqrt(13))/3 ,x=-(4+sqrt(13))/3
Find the zeros of 3x^2+8x=-1
The zeros of 3x^2+8x=-1 are x=(-4+sqrt(13))/3 ,x=-(4+sqrt(13))/3