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

The solutions to the equation 6(2x+4)^2=(2x+4)+2 are x=-5/3 ,x=-9/4

Find the zeros of 6(2x+4)^2=(2x+4)+2

The zeros of 6(2x+4)^2=(2x+4)+2 are x=-5/3 ,x=-9/4

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6 (2 x + 4)^{2} = (2 x + 4) + 2

x = - \frac{5}{3}, x = - \frac{9}{4}

Solution steps

6 (2 x + 4)^{2} = (2 x + 4) + 2

Expand

6 (2 x + 4)^{2} : 24 x^{2} + 96 x + 96

24 x^{2} + 96 x + 96 = 2 x + 6

Move

6

to the left side

24 x^{2} + 96 x + 90 = 2 x

Move

2 x

to the left side

24 x^{2} + 94 x + 90 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 94 \pm 9 4 ^{2} - 4 \cdot 24 \cdot 90}{2 \cdot 24}

9 4^{2} - 4 \cdot 24 \cdot 90 = 14

x_{1, 2} = \frac{- 94 \pm 14}{2 \cdot 24}

Separate the solutions

x_{1} = \frac{- 94 + 14}{2 \cdot 24}, x_{2} = \frac{- 94 - 14}{2 \cdot 24}

x = \frac{- 94 + 14}{2 \cdot 24} : - \frac{5}{3}

x = \frac{- 94 - 14}{2 \cdot 24} : - \frac{9}{4}

The solutions to the quadratic equation are:

x = - \frac{5}{3}, x = - \frac{9}{4}

2 x^{3} - x^{2} + 50 x - 25

(a^{- 2} \cdot 8^{7})^{2}

\frac{2}{5} \cdot \frac{5}{6}

\frac{2 π}{6}

x^{2} - 7 x + 2 = 0

What are the solutions to the equation 6(2x+4)^2=(2x+4)+2 ?
The solutions to the equation 6(2x+4)^2=(2x+4)+2 are x=-5/3 ,x=-9/4
Find the zeros of 6(2x+4)^2=(2x+4)+2
The zeros of 6(2x+4)^2=(2x+4)+2 are x=-5/3 ,x=-9/4