What are the solutions to the equation (x+4)/5-x+4= x/3*(x-2)/2 ?

The solutions to the equation (x+4)/5-x+4= x/3*(x-2)/2 are x=(-7+sqrt(769))/5 ,x=-(7+sqrt(769))/5

Find the zeros of (x+4)/5-x+4= x/3*(x-2)/2

The zeros of (x+4)/5-x+4= x/3*(x-2)/2 are x=(-7+sqrt(769))/5 ,x=-(7+sqrt(769))/5

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\frac{x + 4}{5} - x + 4 = \frac{x}{3} \cdot \frac{x - 2}{2}

x = \frac{- 7 + 769}{5}, x = - \frac{7 + 769}{5}

Decimal Notation

x = 4.14616 \dots, x = - 6.94616 \dots

Solution steps

\frac{x + 4}{5} - x + 4 = \frac{x}{3} \cdot \frac{x - 2}{2}

Simplify

3 \cdot 2 : 6

\frac{x + 4}{5} - x + 4 = \frac{x ( x - 2 )}{6}

Find Least Common Multiplier of

5, 6 : 30

Multiply by LCM=

30

\frac{x + 4}{5} \cdot 30 - x \cdot 30 + 4 \cdot 30 = \frac{x ( x - 2 )}{6} \cdot 30

Simplify

6 (x + 4) - 30 x + 120 = 5 x (x - 2)

Expand

6 (x + 4) - 30 x + 120 : - 24 x + 144

Expand

5 x (x - 2) : 5 x^{2} - 10 x

- 24 x + 144 = 5 x^{2} - 10 x

Switch sides

5 x^{2} - 10 x = - 24 x + 144

Move

144

to the left side

5 x^{2} - 10 x - 144 = - 24 x

Move

24 x

to the left side

5 x^{2} + 14 x - 144 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 5 ( - 144 )}{2 \cdot 5}

1 4^{2} - 4 \cdot 5 (- 144) = 2769

x_{1, 2} = \frac{- 14 \pm 2 769}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 14 + 2 769}{2 \cdot 5}, x_{2} = \frac{- 14 - 2 769}{2 \cdot 5}

x = \frac{- 14 + 2 769}{2 \cdot 5} : \frac{- 7 + 769}{5}

x = \frac{- 14 - 2 769}{2 \cdot 5} : - \frac{7 + 769}{5}

The solutions to the quadratic equation are:

x = \frac{- 7 + 769}{5}, x = - \frac{7 + 769}{5}

2 x^{2} + 12 x = 0

(x^{2} + 2)^{2} - 5 (x^{2} + 2) > 6

- 4 + 8

x^{5} - 5 x^{3} - 6 x

x - 3 \leq 12

What are the solutions to the equation (x+4)/5-x+4= x/3*(x-2)/2 ?
The solutions to the equation (x+4)/5-x+4= x/3*(x-2)/2 are x=(-7+sqrt(769))/5 ,x=-(7+sqrt(769))/5
Find the zeros of (x+4)/5-x+4= x/3*(x-2)/2
The zeros of (x+4)/5-x+4= x/3*(x-2)/2 are x=(-7+sqrt(769))/5 ,x=-(7+sqrt(769))/5