Solve (x+t)^2-x^2=17

$$( x + t ) ^ { 2 } - x ^ { 2 } = 17$$

Solve for x

$$x=-\frac{t}{2}+\frac{17}{2t}$$

$$t\neq 0$$

Steps for Solving Linear Equation

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

$$x^{2}+2xt+t^{2}-x^{2}=17$$

Combine x^{2} and -x^{2} to get 0.

$$2xt+t^{2}=17$$

Subtract t^{2} from both sides.

$$2xt=17-t^{2}$$

The equation is in standard form.

$$2tx=17-t^{2}$$

Divide both sides by 2t.

$$\frac{2tx}{2t}=\frac{17-t^{2}}{2t}$$

Dividing by 2t undoes the multiplication by 2t.

$$x=\frac{17-t^{2}}{2t}$$

Divide 17-t^{2} by 2t.

$$x=-\frac{t}{2}+\frac{17}{2t}$$

Solve for t

$$t=\sqrt{x^{2}+17}-x$$

$$t=-\left(\sqrt{x^{2}+17}+x\right)$$