Solve {l}{f(x)=(m-3)x-(m-2)^2+5text{jest}}{text{ikaztego,ze:}}

$$\left. \begin{array} { l } { f ( x ) = ( m - 3 ) x - ( m - 2 ) ^ { 2 } + 5 \text { jest } } \\ { \text { ika z tego, ze: } } \end{array} \right.$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=-\frac{m^{2}-4m-1}{3+f-m}\text{, }&f\neq m-3\\x\in \mathrm{C}\text{, }&\left(m=\sqrt{5}+2\text{ and }f=\sqrt{5}-1\right)\text{ or }\left(m=2-\sqrt{5}\text{ and }f=-\sqrt{5}-1\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply m-3 by x.

$$fx=mx-3x-\left(m-2\right)^{2}+5$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(m-2\right)^{2}.

$$fx=mx-3x-\left(m^{2}-4m+4\right)+5$$

To find the opposite of m^{2}-4m+4, find the opposite of each term.

$$fx=mx-3x-m^{2}+4m-4+5$$

Add -4 and 5 to get 1.

$$fx=mx-3x-m^{2}+4m+1$$

Subtract mx from both sides.

$$fx-mx=-3x-m^{2}+4m+1$$

Add 3x to both sides.

$$fx-mx+3x=-m^{2}+4m+1$$

Combine all terms containing x.

$$\left(f-m+3\right)x=-m^{2}+4m+1$$

The equation is in standard form.

$$\left(3+f-m\right)x=1+4m-m^{2}$$

Divide both sides by f-m+3.

$$\frac{\left(3+f-m\right)x}{3+f-m}=\frac{1+4m-m^{2}}{3+f-m}$$

Dividing by f-m+3 undoes the multiplication by f-m+3.

$$x=\frac{1+4m-m^{2}}{3+f-m}$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{m^{2}-4m-1}{3+f-m}\text{, }&f\neq m-3\\x\in \mathrm{R}\text{, }&\left(m=2-\sqrt{5}\text{ and }f=-\sqrt{5}-1\right)\text{ or }\left(m=\sqrt{5}+2\text{ and }f=\sqrt{5}-1\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply m-3 by x.

$$fx=mx-3x-\left(m-2\right)^{2}+5$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(m-2\right)^{2}.

$$fx=mx-3x-\left(m^{2}-4m+4\right)+5$$

To find the opposite of m^{2}-4m+4, find the opposite of each term.

$$fx=mx-3x-m^{2}+4m-4+5$$

Add -4 and 5 to get 1.

$$fx=mx-3x-m^{2}+4m+1$$

Subtract mx from both sides.

$$fx-mx=-3x-m^{2}+4m+1$$

Add 3x to both sides.

$$fx-mx+3x=-m^{2}+4m+1$$

Combine all terms containing x.

$$\left(f-m+3\right)x=-m^{2}+4m+1$$

The equation is in standard form.

$$\left(3+f-m\right)x=1+4m-m^{2}$$

Divide both sides by f-m+3.

$$\frac{\left(3+f-m\right)x}{3+f-m}=\frac{1+4m-m^{2}}{3+f-m}$$

Dividing by f-m+3 undoes the multiplication by f-m+3.

$$x=\frac{1+4m-m^{2}}{3+f-m}$$