Solve -1/64^2+4b+c=0

$$- \frac{ 1 }{ 6 } { 4 }^{ 2 } +4b+c = 0$$

Solve for b

$$b=-\frac{c}{4}+\frac{2}{3}$$

Steps for Solving Linear Equation

Calculate 4 to the power of 2 and get 16.

$$-\frac{1}{6}\times 16+4b+c=0$$

Multiply -\frac{1}{6} and 16 to get -\frac{8}{3}.

$$-\frac{8}{3}+4b+c=0$$

Add \frac{8}{3} to both sides. Anything plus zero gives itself.

$$4b+c=\frac{8}{3}$$

Subtract c from both sides.

$$4b=\frac{8}{3}-c$$

Divide both sides by 4.

$$\frac{4b}{4}=\frac{\frac{8}{3}-c}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$b=\frac{\frac{8}{3}-c}{4}$$

Divide \frac{8}{3}-c by 4.

$$b=-\frac{c}{4}+\frac{2}{3}$$

Solve for c

$$c=\frac{8}{3}-4b$$

Solution Steps

Calculate 4 to the power of 2 and get 16.

$$-\frac{1}{6}\times 16+4b+c=0$$

Multiply -\frac{1}{6} and 16 to get -\frac{8}{3}.

$$-\frac{8}{3}+4b+c=0$$

Add \frac{8}{3} to both sides. Anything plus zero gives itself.

$$4b+c=\frac{8}{3}$$

Subtract 4b from both sides.

$$c=\frac{8}{3}-4b$$