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

Mar 26, 2018

${x}_{1} = \frac{2}{3}$ and ${x}_{2} = 2$

#### Explanation:

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

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

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

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

$\left(3 x - 2\right) \cdot \left(x - 2\right) = 0$

Hence ${x}_{1} = \frac{2}{3}$ and ${x}_{2} = 2$