What are the solutions to the equation ((3x)}{2-5}=\frac{(2x(x+3)))/5 ?

The solutions to the equation ((3x)}{2-5}=\frac{(2x(x+3)))/5 are x=0,x=-11/2

Find the zeros of ((3x)}{2-5}=\frac{(2x(x+3)))/5

The zeros of ((3x)}{2-5}=\frac{(2x(x+3)))/5 are x=0,x=-11/2

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

x = 0, x = - \frac{11}{2}

Decimal Notation

x = 0, x = - 5.5

Solution steps

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

Simplify

2 - 5 : - 3

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

Find Least Common Multiplier of

3, 5 : 15

Multiply by LCM=

15

\frac{3 x}{- 3} \cdot 15 = \frac{2 x ( x + 3 )}{5} \cdot 15

Simplify

- 15 x = 6 x (x + 3)

Expand

6 x (x + 3) : 6 x^{2} + 18 x

- 15 x = 6 x^{2} + 18 x

Switch sides

6 x^{2} + 18 x = - 15 x

Move

15 x

to the left side

6 x^{2} + 33 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 33 \pm 3 3 ^{2} - 4 \cdot 6 \cdot 0}{2 \cdot 6}

3 3^{2} - 4 \cdot 6 \cdot 0 = 33

x_{1, 2} = \frac{- 33 \pm 33}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 33 + 33}{2 \cdot 6}, x_{2} = \frac{- 33 - 33}{2 \cdot 6}

x = \frac{- 33 + 33}{2 \cdot 6} : 0

x = \frac{- 33 - 33}{2 \cdot 6} : - \frac{11}{2}

The solutions to the quadratic equation are:

x = 0, x = - \frac{11}{2}

4 x^{2} - 2 (k + 1) x + (k + 4) = 0

x, f = x^{2} + (m - 2) x + (m - 3)

\frac{x ^{2} - 5 x}{6 + 1} = \frac{2 x + 11}{3}

62 \cdot x^{2} + 5 x - 32 = 0

y^{2} - 6 y + 2 = 0

What are the solutions to the equation ((3x)}{2-5}=\frac{(2x(x+3)))/5 ?
The solutions to the equation ((3x)}{2-5}=\frac{(2x(x+3)))/5 are x=0,x=-11/2
Find the zeros of ((3x)}{2-5}=\frac{(2x(x+3)))/5
The zeros of ((3x)}{2-5}=\frac{(2x(x+3)))/5 are x=0,x=-11/2