What are the solutions to the equation (x(3x-1))/2-(6x)/3 =(2x^2)/4+(10-3x^2)/2 ?

The solutions to the equation (x(3x-1))/2-(6x)/3 =(2x^2)/4+(10-3x^2)/2 are x=2,x=-1

Find the zeros of (x(3x-1))/2-(6x)/3 =(2x^2)/4+(10-3x^2)/2

The zeros of (x(3x-1))/2-(6x)/3 =(2x^2)/4+(10-3x^2)/2 are x=2,x=-1

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\frac{x ( 3 x - 1 )}{2} - \frac{6 x}{3} = \frac{2 x ^{2}}{4} + \frac{10 - 3 x ^{2}}{2}

x = 2, x = - 1

Solution steps

\frac{x ( 3 x - 1 )}{2} - \frac{6 x}{3} = \frac{2 x ^{2}}{4} + \frac{10 - 3 x ^{2}}{2}

Find Least Common Multiplier of

2, 3, 4 : 12

Multiply by LCM=

12

\frac{x ( 3 x - 1 )}{2} \cdot 12 - \frac{6 x}{3} \cdot 12 = \frac{2 x ^{2}}{4} \cdot 12 + \frac{10 - 3 x ^{2}}{2} \cdot 12

Simplify

6 x (3 x - 1) - 24 x = 6 x^{2} + 6 (- 3 x^{2} + 10)

Expand

6 x (3 x - 1) - 24 x : 18 x^{2} - 30 x

Expand

6 x^{2} + 6 (- 3 x^{2} + 10) : - 12 x^{2} + 60

18 x^{2} - 30 x = - 12 x^{2} + 60

Move

60

to the left side

18 x^{2} - 30 x - 60 = - 12 x^{2}

Move

12 x^{2}

to the left side

30 x^{2} - 30 x - 60 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 30 ) \pm ( - 30 ) ^{2} - 4 \cdot 30 ( - 60 )}{2 \cdot 30}

(- 30)^{2} - 4 \cdot 30 (- 60) = 90

x_{1, 2} = \frac{- ( - 30 ) \pm 90}{2 \cdot 30}

Separate the solutions

x_{1} = \frac{- ( - 30 ) + 90}{2 \cdot 30}, x_{2} = \frac{- ( - 30 ) - 90}{2 \cdot 30}

x = \frac{- ( - 30 ) + 90}{2 \cdot 30} : 2

x = \frac{- ( - 30 ) - 90}{2 \cdot 30} : - 1

The solutions to the quadratic equation are:

x = 2, x = - 1

9 \cdot 16

- 4 x + 5 \leq - 3

39 \geq 3 + 4 n

\frac{2}{3} a - \frac{3}{4} = \frac{3}{4} a

\frac{1}{5} + \frac{1}{4}

What are the solutions to the equation (x(3x-1))/2-(6x)/3 =(2x^2)/4+(10-3x^2)/2 ?
The solutions to the equation (x(3x-1))/2-(6x)/3 =(2x^2)/4+(10-3x^2)/2 are x=2,x=-1
Find the zeros of (x(3x-1))/2-(6x)/3 =(2x^2)/4+(10-3x^2)/2
The zeros of (x(3x-1))/2-(6x)/3 =(2x^2)/4+(10-3x^2)/2 are x=2,x=-1