What are the solutions to the equation 9x^2+5x=12+6x+3x^2 ?

The solutions to the equation 9x^2+5x=12+6x+3x^2 are x=-4/3 ,x= 3/2

Find the zeros of 9x^2+5x=12+6x+3x^2

The zeros of 9x^2+5x=12+6x+3x^2 are x=-4/3 ,x= 3/2

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9 x^{2} + 5 x = 12 + 6 x + 3 x^{2}

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

Decimal Notation

x = - 1.33333 \dots, x = 1.5

Solution steps

9 x^{2} + 5 x = 12 + 6 x + 3 x^{2}

Switch sides

12 + 6 x + 3 x^{2} = 9 x^{2} + 5 x

Move

5 x

to the left side

3 x^{2} + x + 12 = 9 x^{2}

Move

9 x^{2}

to the left side

- 6 x^{2} + x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 6 ) \cdot 12}{2 ( - 6 )}

1^{2} - 4 (- 6) \cdot 12 = 17

x_{1, 2} = \frac{- 1 \pm 17}{2 ( - 6 )}

Separate the solutions

x_{1} = \frac{- 1 + 17}{2 ( - 6 )}, x_{2} = \frac{- 1 - 17}{2 ( - 6 )}

x = \frac{- 1 + 17}{2 ( - 6 )} : - \frac{4}{3}

x = \frac{- 1 - 17}{2 ( - 6 )} : \frac{3}{2}

The solutions to the quadratic equation are:

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

2 x^{2} - x - 1

16 n^{2} = 25

y, (x + 2)^{2} + (y - 4)^{2} = - 12

240 = 128 t - 16 t^{2}

y, y^{2} = 4 x^{2} + x^{3} - 36 - 9 x

What are the solutions to the equation 9x^2+5x=12+6x+3x^2 ?
The solutions to the equation 9x^2+5x=12+6x+3x^2 are x=-4/3 ,x= 3/2
Find the zeros of 9x^2+5x=12+6x+3x^2
The zeros of 9x^2+5x=12+6x+3x^2 are x=-4/3 ,x= 3/2