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

The solutions to the equation (x(x-1))/2 =(2x^2+3)/3 are x=-3/2-i(sqrt(15))/2 ,x=-3/2+i(sqrt(15))/2

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

The zeros of (x(x-1))/2 =(2x^2+3)/3 are x=-3/2-i(sqrt(15))/2 ,x=-3/2+i(sqrt(15))/2

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

x = - \frac{3}{2} - i \frac{15}{2}, x = - \frac{3}{2} + i \frac{15}{2}

Solution steps

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

Find Least Common Multiplier of

2, 3 : 6

Multiply by LCM=

6

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

Simplify

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

Expand

3 x (x - 1) : 3 x^{2} - 3 x

Expand

2 (2 x^{2} + 3) : 4 x^{2} + 6

3 x^{2} - 3 x = 4 x^{2} + 6

Move

6

to the left side

3 x^{2} - 3 x - 6 = 4 x^{2}

Move

4 x^{2}

to the left side

- x^{2} - 3 x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 1 ) ( - 6 )}{2 ( - 1 )}

Simplify

(- 3)^{2} - 4 (- 1) (- 6) : 15 i

x_{1, 2} = \frac{- ( - 3 ) \pm 15 i}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 15 i}{2 ( - 1 )}, x_{2} = \frac{- ( - 3 ) - 15 i}{2 ( - 1 )}

x = \frac{- ( - 3 ) + 15 i}{2 ( - 1 )} : - \frac{3}{2} - i \frac{15}{2}

x = \frac{- ( - 3 ) - 15 i}{2 ( - 1 )} : - \frac{3}{2} + i \frac{15}{2}

The solutions to the quadratic equation are:

x = - \frac{3}{2} - i \frac{15}{2}, x = - \frac{3}{2} + i \frac{15}{2}

x, x^{2} + y^{3} = 9

6 x^{2} + 2 x = - 5

0 = - 0.6 x^{2} + 3 x + 5

(x + 6) (x - 4) = - 16

(3 - x) (2 - x) - 2 = 0

What are the solutions to the equation (x(x-1))/2 =(2x^2+3)/3 ?
The solutions to the equation (x(x-1))/2 =(2x^2+3)/3 are x=-3/2-i(sqrt(15))/2 ,x=-3/2+i(sqrt(15))/2
Find the zeros of (x(x-1))/2 =(2x^2+3)/3
The zeros of (x(x-1))/2 =(2x^2+3)/3 are x=-3/2-i(sqrt(15))/2 ,x=-3/2+i(sqrt(15))/2