1 вариант прочтения:
[m]\frac{(3x+4)^2}{36}+ 3x(1–x) = \frac{ (x–4)(x+4)}{12} [/m]
Умножим на 36
(3x+4)^ 2+3*36x(1–x)=3(x^2-16)
9x^2+24x+16+108x-108x^2=3x^2-48;
102x^2-132x-64=0
51x^2-66x-32=0
D/4=33^2+51*32=1089+1632=2721
x_(1)=(33-sqrt(2721))/51; x_(2)=(33+sqrt(2721))/51
Второй вариант
[m]\frac{(3x+4)^2}{36+3x(1–x)}=\frac{(x–4)(x+4)}{12}[/m]
Пропорция
12*(3x+4)^2=3(12+x-x^2)*(x^2-16)
4*(3x+4)^2=(12+x-x^2)*(x^2-16)
4*(x^2+24x+16)=12x^2+x^3-x^4-192-16x+16x^2
x^4-x^3-24x^2+112x+256=0
???