How do you solve $3 (x^{2}) - 6 x + 7 = x^{2} + 3 x - x (x + 1) + 3$ $3(x^2) -6x + 7 = x^2 + 3x -x(x+1) + 3$ ?

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Answer 1 · 2018-03-26T18:50:16

$x_{1} = \frac{2}{3}$ $x_1=2/3$ and $x_{2} = 2$ $x_2=2$

$3 x^{2} - 6 x + 7 = x^{2} + 3 x - x \cdot (x + 1) + 3$ $3x^2-6x+7=x^2+3x-x*(x+1)+3$

$3 x^{2} - 6 x + 7 = x^{2} + 3 x - x^{2} - x + 3$ $3x^2-6x+7=x^2+3x-x^2-x+3$

$3 x^{2} - 6 x + 7 = 2 x + 3$ $3x^2-6x+7=2x+3$

$3 x^{2} - 8 x + 4 = 0$ $3x^2-8x+4=0$

$(3 x - 2) \cdot (x - 2) = 0$ $(3x-2)*(x-2)=0$

Hence $x_{1} = \frac{2}{3}$ $x_1=2/3$ and $x_{2} = 2$ $x_2=2$

How do you solve 3(x2)−6x+7=x2+3x−x(x+1)+33(x^2) -6x + 7 = x^2 + 3x -x(x+1) + 3?