What are the solutions to the equation 3x^2+(-6x)+14-2x^2+(-18)+7x=2x-4 ?

The solutions to the equation 3x^2+(-6x)+14-2x^2+(-18)+7x=2x-4 are x=1,x=0

Find the zeros of 3x^2+(-6x)+14-2x^2+(-18)+7x=2x-4

The zeros of 3x^2+(-6x)+14-2x^2+(-18)+7x=2x-4 are x=1,x=0

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3 x^{2} + (- 6 x) + 14 - 2 x^{2} + (- 18) + 7 x = 2 x - 4

x = 1, x = 0

Solution steps

3 x^{2} + (- 6 x) + 14 - 2 x^{2} + (- 18) + 7 x = 2 x - 4

Refine

x^{2} + x - 4 = 2 x - 4

Move

4

to the left side

x^{2} + x = 2 x

Move

2 x

to the left side

x^{2} - x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot 0 = 1

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

Separate the solutions

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

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

x = \frac{- ( - 1 ) - 1}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

x = 1, x = 0

3 r^{2} + 27 r = 12

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

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

4 n^{2} + 8 n + 4 = 0

(3 x + 2) (3 x - 2) = 0

What are the solutions to the equation 3x^2+(-6x)+14-2x^2+(-18)+7x=2x-4 ?
The solutions to the equation 3x^2+(-6x)+14-2x^2+(-18)+7x=2x-4 are x=1,x=0
Find the zeros of 3x^2+(-6x)+14-2x^2+(-18)+7x=2x-4
The zeros of 3x^2+(-6x)+14-2x^2+(-18)+7x=2x-4 are x=1,x=0