Solve {c}{f(x+1)=t+1^3-5(t+1)^2-4(t+1)+20}{[[t+1]=t+1^3-5[t+1]^2-4(t+1)+20}

$$\left. \begin{array} { c } { f ( x + 1 ) = t + 1 ^ { 3 } - 5 ( t + 1 ) ^ { 2 } - 4 ( t + 1 ) + 20 } \\ { [ [ t + 1 ] = t + 1 ^ { 3 } - 5 [ t + 1 ] ^ { 2 } - 4 ( t + 1 ) + 20 } \end{array} \right.$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=-\frac{5t^{2}+13t+f-12}{f}\text{, }&f\neq 0\\x\in \mathrm{C}\text{, }&\left(t=\frac{-\sqrt{409}-13}{10}\text{ or }t=\frac{\sqrt{409}-13}{10}\right)\text{ and }f=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply f by x+1.

$$fx+f=t+1^{3}-5\left(t+1\right)^{2}-4\left(t+1\right)+20$$

Calculate 1 to the power of 3 and get 1.

$$fx+f=t+1-5\left(t+1\right)^{2}-4\left(t+1\right)+20$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(t+1\right)^{2}.

$$fx+f=t+1-5\left(t^{2}+2t+1\right)-4\left(t+1\right)+20$$

Use the distributive property to multiply -5 by t^{2}+2t+1.

$$fx+f=t+1-5t^{2}-10t-5-4\left(t+1\right)+20$$

Combine t and -10t to get -9t.

$$fx+f=-9t+1-5t^{2}-5-4\left(t+1\right)+20$$

Subtract 5 from 1 to get -4.

$$fx+f=-9t-4-5t^{2}-4\left(t+1\right)+20$$

Use the distributive property to multiply -4 by t+1.

$$fx+f=-9t-4-5t^{2}-4t-4+20$$

Combine -9t and -4t to get -13t.

$$fx+f=-13t-4-5t^{2}-4+20$$

Subtract 4 from -4 to get -8.

$$fx+f=-13t-8-5t^{2}+20$$

Add -8 and 20 to get 12.

$$fx+f=-13t+12-5t^{2}$$

Subtract f from both sides.

$$fx=-13t+12-5t^{2}-f$$

The equation is in standard form.

$$fx=12-f-13t-5t^{2}$$

Divide both sides by f.

$$\frac{fx}{f}=\frac{12-f-13t-5t^{2}}{f}$$

Dividing by f undoes the multiplication by f.

$$x=\frac{12-f-13t-5t^{2}}{f}$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{5t^{2}+13t+f-12}{f}\text{, }&f\neq 0\\x\in \mathrm{R}\text{, }&\left(t=\frac{-\sqrt{409}-13}{10}\text{ or }t=\frac{\sqrt{409}-13}{10}\right)\text{ and }f=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply f by x+1.

$$fx+f=t+1^{3}-5\left(t+1\right)^{2}-4\left(t+1\right)+20$$

Calculate 1 to the power of 3 and get 1.

$$fx+f=t+1-5\left(t+1\right)^{2}-4\left(t+1\right)+20$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(t+1\right)^{2}.

$$fx+f=t+1-5\left(t^{2}+2t+1\right)-4\left(t+1\right)+20$$

Use the distributive property to multiply -5 by t^{2}+2t+1.

$$fx+f=t+1-5t^{2}-10t-5-4\left(t+1\right)+20$$

Combine t and -10t to get -9t.

$$fx+f=-9t+1-5t^{2}-5-4\left(t+1\right)+20$$

Subtract 5 from 1 to get -4.

$$fx+f=-9t-4-5t^{2}-4\left(t+1\right)+20$$

Use the distributive property to multiply -4 by t+1.

$$fx+f=-9t-4-5t^{2}-4t-4+20$$

Combine -9t and -4t to get -13t.

$$fx+f=-13t-4-5t^{2}-4+20$$

Subtract 4 from -4 to get -8.

$$fx+f=-13t-8-5t^{2}+20$$

Add -8 and 20 to get 12.

$$fx+f=-13t+12-5t^{2}$$

Subtract f from both sides.

$$fx=-13t+12-5t^{2}-f$$

The equation is in standard form.

$$fx=12-f-13t-5t^{2}$$

Divide both sides by f.

$$\frac{fx}{f}=\frac{12-f-13t-5t^{2}}{f}$$

Dividing by f undoes the multiplication by f.

$$x=\frac{12-f-13t-5t^{2}}{f}$$