What are the solutions to the equation (x+3)(3x-1)=-4 ?

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

Find the zeros of (x+3)(3x-1)=-4

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

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(x + 3) (3 x - 1) = - 4

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

Decimal

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

Solution steps

(x + 3) (3 x - 1) = - 4

Expand

(x + 3) (3 x - 1) : 3 x^{2} + 8 x - 3

3 x^{2} + 8 x - 3 = - 4

Move

4

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}

6 x^{2} - 29 x + 35

- 4 ∣ x + 2 ∣ - 8 = - 56

n^{2} + 7 n + 15 = 5

1 - 5 (- 7 r + 5) \leq - 5 r + 5 - 4

6 x^{2} + 2 x

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