Solve r=sqrt{(1,5)^2+(-2)^2-C}

$$r = \sqrt { ( 1,5 ) ^ { 2 } + ( - 2 ) ^ { 2 } - C }$$

Solve for C

$$C=6,25-r^{2}$$

$$r\geq 0$$

Steps for Solving Radical Equations

Calculate 1,5 to the power of 2 and get 2,25.

$$r=\sqrt{2,25+\left(-2\right)^{2}-C}$$

Calculate -2 to the power of 2 and get 4.

$$r=\sqrt{2,25+4-C}$$

Add 2,25 and 4 to get 6,25.

$$r=\sqrt{6,25-C}$$

Swap sides so that all variable terms are on the left hand side.

$$\sqrt{6,25-C}=r$$

Square both sides of the equation.

$$-C+6,25=r^{2}$$

Subtract 6,25 from both sides of the equation.

$$-C+6,25-6,25=r^{2}-6,25$$

Subtracting 6,25 from itself leaves 0.

$$-C=r^{2}-6,25$$

Subtract 6,25 from r^{2}.

$$-C=r^{2}-\frac{25}{4}$$

Divide both sides by -1.

$$\frac{-C}{-1}=\frac{r^{2}-\frac{25}{4}}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$C=\frac{r^{2}-\frac{25}{4}}{-1}$$

Divide r^{2}-\frac{25}{4} by -1.

$$C=\frac{25}{4}-r^{2}$$

Solve for r

$$r=\sqrt{6,25-C}$$

$$C\leq 6,25$$