What are the solutions to the equation x/2 (3x-5)= 1/2-3x ?

The solutions to the equation x/2 (3x-5)= 1/2-3x are x=(-1+sqrt(13))/6 ,x=(-1-sqrt(13))/6

Find the zeros of x/2 (3x-5)= 1/2-3x

The zeros of x/2 (3x-5)= 1/2-3x are x=(-1+sqrt(13))/6 ,x=(-1-sqrt(13))/6

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\frac{x}{2} (3 x - 5) = \frac{1}{2} - 3 x

x = \frac{- 1 + 13}{6}, x = \frac{- 1 - 13}{6}

Decimal Notation

x = 0.43425 \dots, x = - 0.76759 \dots

Solution steps

\frac{x}{2} (3 x - 5) = \frac{1}{2} - 3 x

Multiply both sides by

2

x (3 x - 5) = 1 - 6 x

Expand

x (3 x - 5) : 3 x^{2} - 5 x

3 x^{2} - 5 x = 1 - 6 x

Move

6 x

to the left side

3 x^{2} + x = 1

Move

1

to the left side

3 x^{2} + x - 1 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 3 (- 1) = 13

x_{1, 2} = \frac{- 1 \pm 13}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 1 + 13}{2 \cdot 3}, x_{2} = \frac{- 1 - 13}{2 \cdot 3}

x = \frac{- 1 + 13}{2 \cdot 3} : \frac{- 1 + 13}{6}

x = \frac{- 1 - 13}{2 \cdot 3} : \frac{- 1 - 13}{6}

The solutions to the quadratic equation are:

x = \frac{- 1 + 13}{6}, x = \frac{- 1 - 13}{6}

a^{1} + 2 a^{2} = - 1, - 3 a^{1} + 4 a^{2} = k, 2 a^{1} - a^{2} = - 7

6 n^{2} - 18 n - 12 = 0

36.12 + 2. 2^{2} = c^{2}

(2 x - 1) (x + 6) = 0

x^{2} - 12 x + 11

What are the solutions to the equation x/2 (3x-5)= 1/2-3x ?
The solutions to the equation x/2 (3x-5)= 1/2-3x are x=(-1+sqrt(13))/6 ,x=(-1-sqrt(13))/6
Find the zeros of x/2 (3x-5)= 1/2-3x
The zeros of x/2 (3x-5)= 1/2-3x are x=(-1+sqrt(13))/6 ,x=(-1-sqrt(13))/6