What are the solutions to the equation 8x^2+8=6-9x ?

The solutions to the equation 8x^2+8=6-9x are x=(-9+sqrt(17))/(16),x=(-9-sqrt(17))/(16)

Find the zeros of 8x^2+8=6-9x

The zeros of 8x^2+8=6-9x are x=(-9+sqrt(17))/(16),x=(-9-sqrt(17))/(16)

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8 x^{2} + 8 = 6 - 9 x

x = \frac{- 9 + 17}{16}, x = \frac{- 9 - 17}{16}

Solution steps

8 x^{2} + 8 = 6 - 9 x

Move

9 x

to the left side

8 x^{2} + 8 + 9 x = 6

Move

6

to the left side

8 x^{2} + 9 x + 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 8 \cdot 2}{2 \cdot 8}

9^{2} - 4 \cdot 8 \cdot 2 = 17

x_{1, 2} = \frac{- 9 \pm 17}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- 9 + 17}{2 \cdot 8}, x_{2} = \frac{- 9 - 17}{2 \cdot 8}

x = \frac{- 9 + 17}{2 \cdot 8} : \frac{- 9 + 17}{16}

x = \frac{- 9 - 17}{2 \cdot 8} : \frac{- 9 - 17}{16}

The solutions to the quadratic equation are:

x = \frac{- 9 + 17}{16}, x = \frac{- 9 - 17}{16}

y^{2} - 6 y + 5 = 0

2 x + 22 = 4 (x + 3)

w (7 w - 10) + 4 = 0

\frac{11}{8} t - (\frac{7}{8} + \frac{3}{8} t) = 0

3 a^{2} + 6 a

What are the solutions to the equation 8x^2+8=6-9x ?
The solutions to the equation 8x^2+8=6-9x are x=(-9+sqrt(17))/(16),x=(-9-sqrt(17))/(16)
Find the zeros of 8x^2+8=6-9x
The zeros of 8x^2+8=6-9x are x=(-9+sqrt(17))/(16),x=(-9-sqrt(17))/(16)