Question

Решите уравнение, даю пятнадцать​

Answer 1

Ответ:

$(2x - ) {}^{2} = 11x - 19 \\ 4x {}^{2} - 12x + 9 = 11x - 19 \\ 4x {}^{2} - 23x + 28 = 0 \\ x(4x - 7) - 4(4x - 7) = 0 \\ (4x - 7)(x - 4) = 0 \\ \\ 4x - 7 = 0 \\ x - 4 = 0 \\ \\ x1 = \frac{7}{4} \\ x2 = 4$

Answer 2

$(2x-3)^{2} =11x-19\\\\4x^{2}-12x+9=11x-19\\\\4x^{2} -23x+28=0\\\\D=(-23)^{2}-4*4*28=529-448=81=9^{2}\\\\x_{1}=\frac{23-9}{8}=1,75\\\\x_{2}=\frac{23+9}{8}=4\\\\Otvet:\boxed{1,75;4}$