What are the solutions to the equation 9x^2=9-5x ?

The solutions to the equation 9x^2=9-5x are x=(-5+sqrt(349))/(18),x=(-5-sqrt(349))/(18)

Find the zeros of 9x^2=9-5x

The zeros of 9x^2=9-5x are x=(-5+sqrt(349))/(18),x=(-5-sqrt(349))/(18)

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9 x^{2} = 9 - 5 x

x = \frac{- 5 + 349}{18}, x = \frac{- 5 - 349}{18}

Decimal Notation

x = 0.76008 \dots, x = - 1.31564 \dots

Solution steps

9 x^{2} = 9 - 5 x

Move

5 x

to the left side

9 x^{2} + 5 x = 9

Move

9

to the left side

9 x^{2} + 5 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 9 ( - 9 )}{2 \cdot 9}

5^{2} - 4 \cdot 9 (- 9) = 349

x_{1, 2} = \frac{- 5 \pm 349}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- 5 + 349}{2 \cdot 9}, x_{2} = \frac{- 5 - 349}{2 \cdot 9}

x = \frac{- 5 + 349}{2 \cdot 9} : \frac{- 5 + 349}{18}

x = \frac{- 5 - 349}{2 \cdot 9} : \frac{- 5 - 349}{18}

The solutions to the quadratic equation are:

x = \frac{- 5 + 349}{18}, x = \frac{- 5 - 349}{18}

- \frac{3}{4} x \geq - \frac{5}{8} + \frac{2}{3} x

6 n + 3 \leq - 9

(\frac{1}{4})^{- \frac{3}{2}}

y, \frac{y - D}{E} = F

\frac{1}{\frac{1}{100} + \frac{1}{100}}

What are the solutions to the equation 9x^2=9-5x ?
The solutions to the equation 9x^2=9-5x are x=(-5+sqrt(349))/(18),x=(-5-sqrt(349))/(18)
Find the zeros of 9x^2=9-5x
The zeros of 9x^2=9-5x are x=(-5+sqrt(349))/(18),x=(-5-sqrt(349))/(18)