Question

x^2-17x+42=0 ришите пж

Answer 1

$x^2-17x+42=0\\\\D=121 \; ,\; \; x_{1,2}= \frac{17\pm 11}{2} \\\\x_1=3\; ,\; \; x_2=14$

Answer 2

x² - 17x + 42 = 0
x² - 3x - 14x + 42 = 0
x • (x - 3) - 14(x - 3) = 0
(x - 3) • (x - 14) = 0
x - 3 = 0
x - 14 = 0
x1 = 3
x2 = 14