5–(7–10)–10–(5–2)
$x^{3}$ - $6x^{2}$ + 9x = 0
$x^{3}$ - $6x^{2}$ + $9^{x}$ = 0
=
\frac{
(1,28\cdot 10^{12}) \cdot e^{0,087t}
}{
(10)^{10}e^{2 \cdot
0,087t} - 4,49\cdot 10^{6} \cdot e^{0,087t} ...
207-(0.7x18)
𝑥
3 − 6𝑥
2 + 9𝑥 = 0
a+b=9
a-c=-2
a-d=-4
b+d=13
c+d=14
h(x)= -x² + 4x + 20
4x^2 + x^2
4x (x)
