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

The solutions to the equation 2/3 x^2+1/4 x=3 are x=(-3+3sqrt(129))/(16),x=(-3-3sqrt(129))/(16)

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

The zeros of 2/3 x^2+1/4 x=3 are x=(-3+3sqrt(129))/(16),x=(-3-3sqrt(129))/(16)

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

x = \frac{- 3 + 3 129}{16}, x = \frac{- 3 - 3 129}{16}

Solution steps

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

Find Least Common Multiplier of

3, 4 : 12

Multiply by LCM=

12

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

Simplify

8 x^{2} + 3 x = 36

Move

36

to the left side

8 x^{2} + 3 x - 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 8 ( - 36 )}{2 \cdot 8}

3^{2} - 4 \cdot 8 (- 36) = 3129

x_{1, 2} = \frac{- 3 \pm 3 129}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- 3 + 3 129}{2 \cdot 8}, x_{2} = \frac{- 3 - 3 129}{2 \cdot 8}

x = \frac{- 3 + 3 129}{2 \cdot 8} : \frac{- 3 + 3 129}{16}

x = \frac{- 3 - 3 129}{2 \cdot 8} : \frac{- 3 - 3 129}{16}

The solutions to the quadratic equation are:

x = \frac{- 3 + 3 129}{16}, x = \frac{- 3 - 3 129}{16}

2 y^{2} + 11 y + 12

(6 + 8 i) (6 - 8 i)

4 k^{2} - 25

\frac{3}{4} + \frac{5}{8}

x^{2} - y^{2}

What are the solutions to the equation 2/3 x^2+1/4 x=3 ?
The solutions to the equation 2/3 x^2+1/4 x=3 are x=(-3+3sqrt(129))/(16),x=(-3-3sqrt(129))/(16)
Find the zeros of 2/3 x^2+1/4 x=3
The zeros of 2/3 x^2+1/4 x=3 are x=(-3+3sqrt(129))/(16),x=(-3-3sqrt(129))/(16)