What are the solutions to the equation 1/12 x^2+1/3 x-4=0 ?

The solutions to the equation 1/12 x^2+1/3 x-4=0 are x=2(sqrt(13)-1),x=-2(1+sqrt(13))

Find the zeros of 1/12 x^2+1/3 x-4=0

The zeros of 1/12 x^2+1/3 x-4=0 are x=2(sqrt(13)-1),x=-2(1+sqrt(13))

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\frac{1}{12} x^{2} + \frac{1}{3} x - 4 = 0

x = 2 (13 - 1), x = - 2 (1 + 13)

Decimal Notation

x = 5.21110 \dots, x = - 9.21110 \dots

Solution steps

\frac{1}{12} x^{2} + \frac{1}{3} x - 4 = 0

Find Least Common Multiplier of

12, 3 : 12

Multiply by LCM=

12

\frac{1}{12} x^{2} \cdot 12 + \frac{1}{3} x \cdot 12 - 4 \cdot 12 = 0 \cdot 12

Simplify

x^{2} + 4 x - 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 48 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 48) = 413

x_{1, 2} = \frac{- 4 \pm 4 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 4 13}{2 \cdot 1}, x_{2} = \frac{- 4 - 4 13}{2 \cdot 1}

x = \frac{- 4 + 4 13}{2 \cdot 1} : 2 (13 - 1)

x = \frac{- 4 - 4 13}{2 \cdot 1} : - 2 (1 + 13)

The solutions to the quadratic equation are:

x = 2 (13 - 1), x = - 2 (1 + 13)

\frac{x}{2} + 4 = 12

32 a + 28

2 b^{2} + 13 b + 20

∣ - 9 + 8 ∣

36 + 5 n = - 3 (n + 4)

What are the solutions to the equation 1/12 x^2+1/3 x-4=0 ?
The solutions to the equation 1/12 x^2+1/3 x-4=0 are x=2(sqrt(13)-1),x=-2(1+sqrt(13))
Find the zeros of 1/12 x^2+1/3 x-4=0
The zeros of 1/12 x^2+1/3 x-4=0 are x=2(sqrt(13)-1),x=-2(1+sqrt(13))