Solve (v-10)^2-49=0

$$( v - 10 ) ^ { 2 } - 49 = 0$$

Solve for v

$$v=17$$

$$v=3$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(v-10\right)^{2}.

$$v^{2}-20v+100-49=0$$

Subtract 49 from 100 to get 51.

$$v^{2}-20v+51=0$$

To solve the equation, factor v^{2}-20v+51 using formula v^{2}+\left(a+b\right)v+ab=\left(v+a\right)\left(v+b\right). To find a and b, set up a system to be solved.

$$a+b=-20 ab=51$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 51.

$$-1,-51 -3,-17$$

Calculate the sum for each pair.

$$-1-51=-52 -3-17=-20$$

The solution is the pair that gives sum -20.

$$a=-17 b=-3$$

Rewrite factored expression \left(v+a\right)\left(v+b\right) using the obtained values.

$$\left(v-17\right)\left(v-3\right)$$

To find equation solutions, solve v-17=0 and v-3=0.

$$v=17 v=3$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(v-10\right)^{2}.

$$v^{2}-20v+100-49=0$$

Subtract 49 from 100 to get 51.

$$v^{2}-20v+51=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as v^{2}+av+bv+51. To find a and b, set up a system to be solved.

$$a+b=-20 ab=1\times 51=51$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 51.

$$-1,-51 -3,-17$$

Calculate the sum for each pair.

$$-1-51=-52 -3-17=-20$$

The solution is the pair that gives sum -20.

$$a=-17 b=-3$$

Rewrite v^{2}-20v+51 as \left(v^{2}-17v\right)+\left(-3v+51\right).

$$\left(v^{2}-17v\right)+\left(-3v+51\right)$$

Factor out v in the first and -3 in the second group.

$$v\left(v-17\right)-3\left(v-17\right)$$

Factor out common term v-17 by using distributive property.

$$\left(v-17\right)\left(v-3\right)$$

To find equation solutions, solve v-17=0 and v-3=0.

$$v=17 v=3$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(v-10\right)^{2}.

$$v^{2}-20v+100-49=0$$

Subtract 49 from 100 to get 51.

$$v^{2}-20v+51=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -20 for b, and 51 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$v=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 51}}{2}$$

Square -20.

$$v=\frac{-\left(-20\right)±\sqrt{400-4\times 51}}{2}$$

Multiply -4 times 51.

$$v=\frac{-\left(-20\right)±\sqrt{400-204}}{2}$$

Add 400 to -204.

$$v=\frac{-\left(-20\right)±\sqrt{196}}{2}$$

Take the square root of 196.

$$v=\frac{-\left(-20\right)±14}{2}$$

The opposite of -20 is 20.

$$v=\frac{20±14}{2}$$

Now solve the equation v=\frac{20±14}{2} when ± is plus. Add 20 to 14.

$$v=\frac{34}{2}$$

Divide 34 by 2.

$$v=17$$

Now solve the equation v=\frac{20±14}{2} when ± is minus. Subtract 14 from 20.

$$v=\frac{6}{2}$$

Divide 6 by 2.

$$v=3$$

The equation is now solved.

$$v=17 v=3$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(v-10\right)^{2}.

$$v^{2}-20v+100-49=0$$

Subtract 49 from 100 to get 51.

$$v^{2}-20v+51=0$$

Subtract 51 from both sides. Anything subtracted from zero gives its negation.

$$v^{2}-20v=-51$$

Divide -20, the coefficient of the x term, by 2 to get -10. Then add the square of -10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$v^{2}-20v+\left(-10\right)^{2}=-51+\left(-10\right)^{2}$$

Square -10.

$$v^{2}-20v+100=-51+100$$

Add -51 to 100.

$$v^{2}-20v+100=49$$

Factor v^{2}-20v+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(v-10\right)^{2}=49$$

Take the square root of both sides of the equation.

$$\sqrt{\left(v-10\right)^{2}}=\sqrt{49}$$

Simplify.

$$v-10=7 v-10=-7$$

Add 10 to both sides of the equation.

$$v=17 v=3$$