What are the solutions to the equation (3x^2+8x+8}{24}-\frac{x(x-3))/4 =x ?

The solutions to the equation (3x^2+8x+8}{24}-\frac{x(x-3))/4 =x are x=-4/3 ,x=2

Find the zeros of (3x^2+8x+8}{24}-\frac{x(x-3))/4 =x

The zeros of (3x^2+8x+8}{24}-\frac{x(x-3))/4 =x are x=-4/3 ,x=2

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\frac{3 x ^{2} + 8 x + 8}{24} - \frac{x ( x - 3 )}{4} = x

x = - \frac{4}{3}, x = 2

Decimal Notation

x = - 1.33333 \dots, x = 2

Solution steps

\frac{3 x ^{2} + 8 x + 8}{24} - \frac{x ( x - 3 )}{4} = x

Find Least Common Multiplier of

24, 4 : 24

Multiply by LCM=

24

\frac{3 x ^{2} + 8 x + 8}{24} \cdot 24 - \frac{x ( x - 3 )}{4} \cdot 24 = x \cdot 24

Simplify

3 x^{2} + 8 x + 8 - 6 x (x - 3) = 24 x

Expand

3 x^{2} + 8 x + 8 - 6 x (x - 3) : - 3 x^{2} + 26 x + 8

- 3 x^{2} + 26 x + 8 = 24 x

Move

24 x

to the left side

- 3 x^{2} + 2 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 3 ) \cdot 8}{2 ( - 3 )}

2^{2} - 4 (- 3) \cdot 8 = 10

x_{1, 2} = \frac{- 2 \pm 10}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 2 + 10}{2 ( - 3 )}, x_{2} = \frac{- 2 - 10}{2 ( - 3 )}

x = \frac{- 2 + 10}{2 ( - 3 )} : - \frac{4}{3}

x = \frac{- 2 - 10}{2 ( - 3 )} : 2

The solutions to the quadratic equation are:

x = - \frac{4}{3}, x = 2

4 u^{2} = 8 u

1 0^{2} + x^{2} = 1 4^{2}

2 y^{2} + y - 1 = 0, 2 sin^{2} (x) + sin (x) - 1 = 0

(- x - 2)^{2} + x^{2} + 6 (- x - 2) - 6 x + 14 = 0

5 y^{2} + 24 y = 0

What are the solutions to the equation (3x^2+8x+8}{24}-\frac{x(x-3))/4 =x ?
The solutions to the equation (3x^2+8x+8}{24}-\frac{x(x-3))/4 =x are x=-4/3 ,x=2
Find the zeros of (3x^2+8x+8}{24}-\frac{x(x-3))/4 =x
The zeros of (3x^2+8x+8}{24}-\frac{x(x-3))/4 =x are x=-4/3 ,x=2