Solve 5s^2+(17-5s)^2=49

$$5 s ^ { 2 } + ( 17 - 5 s ) ^ { 2 } = 49$$

Solve for s

$$s=3$$

$$s = \frac{8}{3} = 2\frac{2}{3} \approx 2.666666667$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(17-5s\right)^{2}.

$$5s^{2}+289-170s+25s^{2}=49$$

Combine 5s^{2} and 25s^{2} to get 30s^{2}.

$$30s^{2}+289-170s=49$$

Subtract 49 from both sides.

$$30s^{2}+289-170s-49=0$$

Subtract 49 from 289 to get 240.

$$30s^{2}+240-170s=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$30s^{2}-170s+240=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 30 for a, -170 for b, and 240 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$s=\frac{-\left(-170\right)±\sqrt{\left(-170\right)^{2}-4\times 30\times 240}}{2\times 30}$$

Square -170.

$$s=\frac{-\left(-170\right)±\sqrt{28900-4\times 30\times 240}}{2\times 30}$$

Multiply -4 times 30.

$$s=\frac{-\left(-170\right)±\sqrt{28900-120\times 240}}{2\times 30}$$

Multiply -120 times 240.

$$s=\frac{-\left(-170\right)±\sqrt{28900-28800}}{2\times 30}$$

Add 28900 to -28800.

$$s=\frac{-\left(-170\right)±\sqrt{100}}{2\times 30}$$

Take the square root of 100.

$$s=\frac{-\left(-170\right)±10}{2\times 30}$$

The opposite of -170 is 170.

$$s=\frac{170±10}{2\times 30}$$

Multiply 2 times 30.

$$s=\frac{170±10}{60}$$

Now solve the equation s=\frac{170±10}{60} when ± is plus. Add 170 to 10.

$$s=\frac{180}{60}$$

Divide 180 by 60.

$$s=3$$

Now solve the equation s=\frac{170±10}{60} when ± is minus. Subtract 10 from 170.

$$s=\frac{160}{60}$$

Reduce the fraction \frac{160}{60} to lowest terms by extracting and canceling out 20.

$$s=\frac{8}{3}$$

The equation is now solved.

$$s=3 s=\frac{8}{3}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(17-5s\right)^{2}.

$$5s^{2}+289-170s+25s^{2}=49$$

Combine 5s^{2} and 25s^{2} to get 30s^{2}.

$$30s^{2}+289-170s=49$$

Subtract 289 from both sides.

$$30s^{2}-170s=49-289$$

Subtract 289 from 49 to get -240.

$$30s^{2}-170s=-240$$

Divide both sides by 30.

$$\frac{30s^{2}-170s}{30}=-\frac{240}{30}$$

Dividing by 30 undoes the multiplication by 30.

$$s^{2}+\left(-\frac{170}{30}\right)s=-\frac{240}{30}$$

Reduce the fraction \frac{-170}{30} to lowest terms by extracting and canceling out 10.

$$s^{2}-\frac{17}{3}s=-\frac{240}{30}$$

Divide -240 by 30.

$$s^{2}-\frac{17}{3}s=-8$$

Divide -\frac{17}{3}, the coefficient of the x term, by 2 to get -\frac{17}{6}. Then add the square of -\frac{17}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$s^{2}-\frac{17}{3}s+\left(-\frac{17}{6}\right)^{2}=-8+\left(-\frac{17}{6}\right)^{2}$$

Square -\frac{17}{6} by squaring both the numerator and the denominator of the fraction.

$$s^{2}-\frac{17}{3}s+\frac{289}{36}=-8+\frac{289}{36}$$

Add -8 to \frac{289}{36}.

$$s^{2}-\frac{17}{3}s+\frac{289}{36}=\frac{1}{36}$$

Factor s^{2}-\frac{17}{3}s+\frac{289}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(s-\frac{17}{6}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(s-\frac{17}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$s-\frac{17}{6}=\frac{1}{6} s-\frac{17}{6}=-\frac{1}{6}$$

Add \frac{17}{6} to both sides of the equation.

$$s=3 s=\frac{8}{3}$$