What are the solutions to the equation (x^2)/3-(3x)/4 =(1-x)/6 ?

The solutions to the equation (x^2)/3-(3x)/4 =(1-x)/6 are x=2,x=-1/4

Find the zeros of (x^2)/3-(3x)/4 =(1-x)/6

The zeros of (x^2)/3-(3x)/4 =(1-x)/6 are x=2,x=-1/4

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\frac{x ^{2}}{3} - \frac{3 x}{4} = \frac{1 - x}{6}

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

Decimal Notation

x = 2, x = - 0.25

Solution steps

\frac{x ^{2}}{3} - \frac{3 x}{4} = \frac{1 - x}{6}

Find Least Common Multiplier of

3, 4, 6 : 12

Multiply by LCM=

12

\frac{x ^{2}}{3} \cdot 12 - \frac{3 x}{4} \cdot 12 = \frac{1 - x}{6} \cdot 12

Simplify

4 x^{2} - 9 x = 2 (- x + 1)

Expand

2 (- x + 1) : - 2 x + 2

4 x^{2} - 9 x = - 2 x + 2

Move

2

to the left side

4 x^{2} - 9 x - 2 = - 2 x

Move

2 x

to the left side

4 x^{2} - 7 x - 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 4 ( - 2 )}{2 \cdot 4}

(- 7)^{2} - 4 \cdot 4 (- 2) = 9

x_{1, 2} = \frac{- ( - 7 ) \pm 9}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 9}{2 \cdot 4}, x_{2} = \frac{- ( - 7 ) - 9}{2 \cdot 4}

x = \frac{- ( - 7 ) + 9}{2 \cdot 4} : 2

x = \frac{- ( - 7 ) - 9}{2 \cdot 4} : - \frac{1}{4}

The solutions to the quadratic equation are:

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

1653.75 = 1500 (1 + 2 x + x^{2})

x, x^{2} + 5 y^{2} + e^{2 y} = 5

(x + 4)^{2} - 5 = 0

(x - 8)^{2} = 10

5 x^{2} + 10 x = 0

What are the solutions to the equation (x^2)/3-(3x)/4 =(1-x)/6 ?
The solutions to the equation (x^2)/3-(3x)/4 =(1-x)/6 are x=2,x=-1/4
Find the zeros of (x^2)/3-(3x)/4 =(1-x)/6
The zeros of (x^2)/3-(3x)/4 =(1-x)/6 are x=2,x=-1/4