Решите через дискриминантx^2-7x-18<0-6x^2+13x-5≤0x^2-6x-7=0

1.
$x^{2} -7x-18\ \textless \ 0 \\ \\ D=(-7)^2-4*(-18)=121\ \textgreater \ 0 \\ \\ x1= \frac{7+11}{2}=9 \\ x2= \frac{7-11}{2}=-2$
(-2; 9)
2.
$-6x^2+13x-5 \leq 0 \\ D=13^2-4*(-6)*(-5)=49\ \textgreater \ 0 \\ x1= \frac{-13+7}{2*(-6)} =0.5 \\ x2= \frac{-13-7}{2*(-6)}=1 \frac{2}{3}$
$[0.5; 1 \frac{2}{3}]$
3.
$x^{2} -6x-7=0 \\ D=-6^2-4*1*(-7)=64\ \textgreater \ 0 \\ x1= \frac{6+8}{2} =7 \\ x2= \frac{6-8}{2} =-1 \\$
Ответ: -1, 7