Solve sqrt{1-3x+5/4^2}=16

$$\sqrt{ \frac{ 1-3x+5 }{ { 4 }^{ 2 } } } =16$$

Solve for x

$$x = -\frac{4090}{3} = -1363\frac{1}{3} \approx -1363.333333333$$

Steps for Solving Radical Equations

Add 1 and 5 to get 6.

$$\sqrt{\frac{6-3x}{4^{2}}}=16$$

Calculate 4 to the power of 2 and get 16.

$$\sqrt{\frac{6-3x}{16}}=16$$

Divide each term of 6-3x by 16 to get \frac{3}{8}-\frac{3}{16}x.

$$\sqrt{\frac{3}{8}-\frac{3}{16}x}=16$$

Square both sides of the equation.

$$-\frac{3}{16}x+\frac{3}{8}=256$$

Subtract \frac{3}{8} from both sides of the equation.

$$-\frac{3}{16}x+\frac{3}{8}-\frac{3}{8}=256-\frac{3}{8}$$

Subtracting \frac{3}{8} from itself leaves 0.

$$-\frac{3}{16}x=256-\frac{3}{8}$$

Subtract \frac{3}{8} from 256.

$$-\frac{3}{16}x=\frac{2045}{8}$$

Divide both sides of the equation by -\frac{3}{16}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{3}{16}x}{-\frac{3}{16}}=\frac{\frac{2045}{8}}{-\frac{3}{16}}$$

Dividing by -\frac{3}{16} undoes the multiplication by -\frac{3}{16}.

$$x=\frac{\frac{2045}{8}}{-\frac{3}{16}}$$

Divide \frac{2045}{8} by -\frac{3}{16} by multiplying \frac{2045}{8} by the reciprocal of -\frac{3}{16}.

$$x=-\frac{4090}{3}$$