What are the solutions to the equation ((x+2)(x-2))/4-((x-3)^2)/3 =(x(11-x))/6 ?

The solutions to the equation ((x+2)(x-2))/4-((x-3)^2)/3 =(x(11-x))/6 are x=6,x=-8

Find the zeros of ((x+2)(x-2))/4-((x-3)^2)/3 =(x(11-x))/6

The zeros of ((x+2)(x-2))/4-((x-3)^2)/3 =(x(11-x))/6 are x=6,x=-8

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\frac{( x + 2 ) ( x - 2 )}{4} - \frac{( x - 3 ) ^{2}}{3} = \frac{x ( 11 - x )}{6}

x = 6, x = - 8

Solution steps

\frac{( x + 2 ) ( x - 2 )}{4} - \frac{( x - 3 ) ^{2}}{3} = \frac{x ( 11 - x )}{6}

Find Least Common Multiplier of

4, 3, 6 : 12

Multiply by LCM=

12

\frac{( x + 2 ) ( x - 2 )}{4} \cdot 12 - \frac{( x - 3 ) ^{2}}{3} \cdot 12 = \frac{x ( 11 - x )}{6} \cdot 12

Simplify

3 (x + 2) (x - 2) - 4 (x - 3)^{2} = 2 x (- x + 11)

Expand

3 (x + 2) (x - 2) - 4 (x - 3)^{2} : - x^{2} + 24 x - 48

Expand

2 x (- x + 11) : - 2 x^{2} + 22 x

- x^{2} + 24 x - 48 = - 2 x^{2} + 22 x

Move

22 x

to the left side

- x^{2} + 2 x - 48 = - 2 x^{2}

Move

2 x^{2}

to the left side

x^{2} + 2 x - 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 48 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 48) = 14

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

Separate the solutions

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

x = \frac{- 2 + 14}{2 \cdot 1} : 6

x = \frac{- 2 - 14}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

x = 6, x = - 8

x^{2} - 6 x - 6 \geq - 1 - 2 x

a \frac{2}{5} b^{3} \cdot a \frac{2}{5} b^{3}

144 - 9 t^{2}

28 - \frac{x}{2} = \frac{x}{5}

x^{2} - 16 = 8

What are the solutions to the equation ((x+2)(x-2))/4-((x-3)^2)/3 =(x(11-x))/6 ?
The solutions to the equation ((x+2)(x-2))/4-((x-3)^2)/3 =(x(11-x))/6 are x=6,x=-8
Find the zeros of ((x+2)(x-2))/4-((x-3)^2)/3 =(x(11-x))/6
The zeros of ((x+2)(x-2))/4-((x-3)^2)/3 =(x(11-x))/6 are x=6,x=-8