Solve (3s-7)^2-49=0

$$( 3 s - 7 ) ^ { 2 } - 49 = 0$$

Solve for s

$$s = \frac{14}{3} = 4\frac{2}{3} \approx 4.666666667$$

$$s=0$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3s-7\right)^{2}.

$$9s^{2}-42s+49-49=0$$

Subtract 49 from 49 to get 0.

$$9s^{2}-42s=0$$

Factor out s.

$$s\left(9s-42\right)=0$$

To find equation solutions, solve s=0 and 9s-42=0.

$$s=0 s=\frac{14}{3}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3s-7\right)^{2}.

$$9s^{2}-42s+49-49=0$$

Subtract 49 from 49 to get 0.

$$9s^{2}-42s=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -42 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$s=\frac{-\left(-42\right)±\sqrt{\left(-42\right)^{2}}}{2\times 9}$$

Take the square root of \left(-42\right)^{2}.

$$s=\frac{-\left(-42\right)±42}{2\times 9}$$

The opposite of -42 is 42.

$$s=\frac{42±42}{2\times 9}$$

Multiply 2 times 9.

$$s=\frac{42±42}{18}$$

Now solve the equation s=\frac{42±42}{18} when ± is plus. Add 42 to 42.

$$s=\frac{84}{18}$$

Reduce the fraction \frac{84}{18} to lowest terms by extracting and canceling out 6.

$$s=\frac{14}{3}$$

Now solve the equation s=\frac{42±42}{18} when ± is minus. Subtract 42 from 42.

$$s=\frac{0}{18}$$

Divide 0 by 18.

$$s=0$$

The equation is now solved.

$$s=\frac{14}{3} s=0$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3s-7\right)^{2}.

$$9s^{2}-42s+49-49=0$$

Subtract 49 from 49 to get 0.

$$9s^{2}-42s=0$$

Divide both sides by 9.

$$\frac{9s^{2}-42s}{9}=\frac{0}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$s^{2}+\left(-\frac{42}{9}\right)s=\frac{0}{9}$$

Reduce the fraction \frac{-42}{9} to lowest terms by extracting and canceling out 3.

$$s^{2}-\frac{14}{3}s=\frac{0}{9}$$

Divide 0 by 9.

$$s^{2}-\frac{14}{3}s=0$$

Divide -\frac{14}{3}, the coefficient of the x term, by 2 to get -\frac{7}{3}. Then add the square of -\frac{7}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$s^{2}-\frac{14}{3}s+\left(-\frac{7}{3}\right)^{2}=\left(-\frac{7}{3}\right)^{2}$$

Square -\frac{7}{3} by squaring both the numerator and the denominator of the fraction.

$$s^{2}-\frac{14}{3}s+\frac{49}{9}=\frac{49}{9}$$

Factor s^{2}-\frac{14}{3}s+\frac{49}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(s-\frac{7}{3}\right)^{2}=\frac{49}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(s-\frac{7}{3}\right)^{2}}=\sqrt{\frac{49}{9}}$$

Simplify.

$$s-\frac{7}{3}=\frac{7}{3} s-\frac{7}{3}=-\frac{7}{3}$$

Add \frac{7}{3} to both sides of the equation.

$$s=\frac{14}{3} s=0$$