What are the solutions to the equation -6-3x+3x^2-4x-8-9x-4=-2x^2+3x^2 ?

The solutions to the equation -6-3x+3x^2-4x-8-9x-4=-2x^2+3x^2 are x=9,x=-1

Find the zeros of -6-3x+3x^2-4x-8-9x-4=-2x^2+3x^2

The zeros of -6-3x+3x^2-4x-8-9x-4=-2x^2+3x^2 are x=9,x=-1

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- 6 - 3 x + 3 x^{2} - 4 x - 8 - 9 x - 4 = - 2 x^{2} + 3 x^{2}

x = 9, x = - 1

Solution steps

- 6 - 3 x + 3 x^{2} - 4 x - 8 - 9 x - 4 = - 2 x^{2} + 3 x^{2}

Refine

3 x^{2} - 16 x - 18 = x^{2}

Move

x^{2}

to the left side

2 x^{2} - 16 x - 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 2 ( - 18 )}{2 \cdot 2}

(- 16)^{2} - 4 \cdot 2 (- 18) = 20

x_{1, 2} = \frac{- ( - 16 ) \pm 20}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 20}{2 \cdot 2}, x_{2} = \frac{- ( - 16 ) - 20}{2 \cdot 2}

x = \frac{- ( - 16 ) + 20}{2 \cdot 2} : 9

x = \frac{- ( - 16 ) - 20}{2 \cdot 2} : - 1

The solutions to the quadratic equation are:

x = 9, x = - 1

x^{2} - 9 x + 39 = 0

v^{2} - 4 v + 3 = 0

x^{2} - x = \frac{3}{2}

- x^{2} - 20 x - 100 = 0

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

What are the solutions to the equation -6-3x+3x^2-4x-8-9x-4=-2x^2+3x^2 ?
The solutions to the equation -6-3x+3x^2-4x-8-9x-4=-2x^2+3x^2 are x=9,x=-1
Find the zeros of -6-3x+3x^2-4x-8-9x-4=-2x^2+3x^2
The zeros of -6-3x+3x^2-4x-8-9x-4=-2x^2+3x^2 are x=9,x=-1