[tex]f\left(x^2-3x\right)+f=x^2-4x[/tex]
[tex]f\left(x^2-3x\right)+f = f\left(x^2-3x\right) =fx^2-f\cdot \:3x =x^2f-3xf=x^2f-3xf+f[/tex]
[tex]x^2f-3xf+f=x^2-4x[/tex]
[tex]x^2f-3xf+f+4x=x^2-4x+4x[/tex]
[tex]x^2f-3xf+f+4x=x^2[/tex]
[tex]x^2f-3xf+f+4x-x^2=x^2-x^2[/tex]
[tex]\left(f-1\right)x^2+\left(-3f+4\right)x+f=0[/tex]
[tex]\frac{-\left(-3f+4\right)+\sqrt{\left(-3f+4\right)^2-4\left(f-1\right)f}}{2\left(f-1\right)}[/tex]
[tex]\left(-3f+4\right)^2-4\left(f-1\right)f =\left(-3f+4\right)^2-4f\left(f-1\right)[/tex]
[tex]\left(-3f\right)^2+2\left(-3f\right)\cdot \:4+4^2 \\\left(-3f\right)^2=9f^2\\2\cdot \:3f\cdot \:4=24f\\4^2=16\\=9f^2-24f+16[/tex]
[tex]=9f^2-24f+16-4\left(f-1\right)f =9f^2-4f^2-24f+4f+16 =5f^2-24f+4f+16 =5f^2-20f+16\\[/tex]
[tex]=\frac{-\left(-3f+4\right)+\sqrt{5f^2-20f+16}}{2\left(f-1\right)}[/tex]
[tex]-\left(-3f+4\right)=\quad 3f-4[/tex]
[tex]\frac{-\left(-3f+4\right)-\sqrt{\left(-3f+4\right)^2-4\left(f-1\right)f}}{2\left(f-1\right)}[/tex]
[tex]-\left(-3f+4\right)=\quad 3f-4[/tex]
[tex]=\frac{3f-4-\sqrt{5f^2-20f+16}}{2\left(f-1\right)}[/tex]
[tex]x=\frac{3f-4+\sqrt{5f^2-20f+16}}{2\left(f-1\right)},\:x=\frac{3f-4-\sqrt{5f^2-20f+16}}{2\left(f-1\right)};\quad \:f\ne \:1[/tex]
Sper că te-am ajutat!
Succes!
