What are the solutions to the equation 3x^2+12x-(x+4)=0 ?

The solutions to the equation 3x^2+12x-(x+4)=0 are x= 1/3 ,x=-4

Find the zeros of 3x^2+12x-(x+4)=0

The zeros of 3x^2+12x-(x+4)=0 are x= 1/3 ,x=-4

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3 x^{2} + 12 x - (x + 4) = 0

x = \frac{1}{3}, x = - 4

Decimal Notation

x = 0.33333 \dots, x = - 4

Solution steps

3 x^{2} + 12 x - (x + 4) = 0

Expand

3 x^{2} + 12 x - (x + 4) : 3 x^{2} + 11 x - 4

3 x^{2} + 11 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 3 ( - 4 )}{2 \cdot 3}

1 1^{2} - 4 \cdot 3 (- 4) = 13

x_{1, 2} = \frac{- 11 \pm 13}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 11 + 13}{2 \cdot 3}, x_{2} = \frac{- 11 - 13}{2 \cdot 3}

x = \frac{- 11 + 13}{2 \cdot 3} : \frac{1}{3}

x = \frac{- 11 - 13}{2 \cdot 3} : - 4

The solutions to the quadratic equation are:

x = \frac{1}{3}, x = - 4

\frac{9 π}{6}

12 y^{4} + 48 y^{5} - 24 y^{6}

\frac{5}{9} - 5

3.46 - g > 9.51

- 2 x + 6 - 3 x \geq - 8 + 4

What are the solutions to the equation 3x^2+12x-(x+4)=0 ?
The solutions to the equation 3x^2+12x-(x+4)=0 are x= 1/3 ,x=-4
Find the zeros of 3x^2+12x-(x+4)=0
The zeros of 3x^2+12x-(x+4)=0 are x= 1/3 ,x=-4