Question

(a-5)(3a+2)
Решите пожалуйста))))

Answer 1

$3 {a}^{2} + 2a - 15a - 10$
$3 {a}^{2} - 13a - 10 = 0$
$d = {b}^{2} - 4ac$
$d = 289$
$\sqrt{d} = \sqrt{289} = 17$
$a1 = \frac{13 + 17}{6} = 5$
$a2 = \frac{13 - 17}{6} = - \frac{2}{3}$

Answer 2

(a-5)(3a+2)=3a²-15a+2a-10=3a²-13a-10

проверка
√D=√169+120=√289=17
a1=1/6[13+17]=5
a2=1/6[13-17]=-2/3