Алгебра срочно..........

Решите задачу:

$2x^2-8x+3=2\left(x^2-4x\right)+3=2\left(x^2-2\cdot2\cdot x+2^2-2^2\right)+3=\\=2\left(\left(x-2\right)^2-4\right)+3=2\left(x-2\right)^2-8+3=2\left(x-2\right)^2-5$

$-3x^2+4x-1=-3\left(x^2-\dfrac{4}{3}x\right)-1=\\\\=-3\left(x^2-2\cdot\dfrac{2}{3}\cdot x+\left(\dfrac{2}{3}\right)^2-\left(\dfrac{2}{3}\right)^2\right)-1=\\\\=-3\left(\left(x-\dfrac{2}{3}\right)^2-\dfrac{4}{9}\right)-1=\\\\=-3\left(x-\dfrac{2}{3}\right)^2+\dfrac{4}{3}-1=-3\left(x-\dfrac{2}{3}\right)^2+\dfrac{1}{3}$