Solve t^2+10x+20=0

$$t ^ { 2 } + 10 x + 20 = 0$$

Solve for x

$$x=-\frac{t^{2}}{10}-2$$

Steps for Solving Linear Equation

Subtract t^{2} from both sides. Anything subtracted from zero gives its negation.

$$10x+20=-t^{2}$$

Subtract 20 from both sides.

$$10x=-t^{2}-20$$

Divide both sides by 10.

$$\frac{10x}{10}=\frac{-t^{2}-20}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x=\frac{-t^{2}-20}{10}$$

Divide -t^{2}-20 by 10.

$$x=-\frac{t^{2}}{10}-2$$

Solve for t (complex solution)

$$t=-\sqrt{-10x-20}$$

$$t=\sqrt{-10x-20}$$

Solve for t

$$t=\sqrt{-10x-20}$$

$$t=-\sqrt{-10x-20}\text{, }x\leq -2$$