|4x+5|=3|\( ^{2} \)
16\(x ^{2} \)+40x+25=9
16\(x ^{2} \)+40x=-16|:16
\(x ^{2} \)+\( \frac{10}{4} \) x=-1
(x+\( \frac{5}{4} \))^2=-1+\( \frac{25}{16} \)=\( \frac{9}{16} \)|\( \sqrt{} \)
1.)x+\( \frac{5}{4} \)=\( \frac{3}{4} \)
x₁=-\( \frac{5}{4} \)+\( \frac{3}{4} \)=-0,5 Supremum
2.)x+\( \frac{5}{4} \)=-\( \frac{3}{4} \)
x₂=-\( \frac{5}{4} \)-\( \frac{3}{4} \)=-2 Infimum