Solve x^2-20x+49=-2

$${ x }^{ 2 } -20x+49=-2$$

Solve for x

$$x=3$$

$$x=17$$

Steps Using Factoring

Add 2 to both sides.

$$x^{2}-20x+49+2=0$$

Add 49 and 2 to get 51.

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

To solve the equation, factor x^{2}-20x+51 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+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(x+a\right)\left(x+b\right) using the obtained values.

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

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

$$x=17 x=3$$

Steps Using Factoring By Grouping

Add 2 to both sides.

$$x^{2}-20x+49+2=0$$

Add 49 and 2 to get 51.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+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 x^{2}-20x+51 as \left(x^{2}-17x\right)+\left(-3x+51\right).

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

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

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

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

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

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

$$x=17 x=3$$

Steps Using the Quadratic Formula

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.

$$x^{2}-20x+49=-2$$

Add 2 to both sides of the equation.

$$x^{2}-20x+49-\left(-2\right)=-2-\left(-2\right)$$

Subtracting -2 from itself leaves 0.

$$x^{2}-20x+49-\left(-2\right)=0$$

Subtract -2 from 49.

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

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}.

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

Square -20.

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

Multiply -4 times 51.

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

Add 400 to -204.

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

Take the square root of 196.

$$x=\frac{-\left(-20\right)±14}{2}$$

The opposite of -20 is 20.

$$x=\frac{20±14}{2}$$

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

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

Divide 34 by 2.

$$x=17$$

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

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

Divide 6 by 2.

$$x=3$$

The equation is now solved.

$$x=17 x=3$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$x^{2}-20x+49=-2$$

Subtract 49 from both sides of the equation.

$$x^{2}-20x+49-49=-2-49$$

Subtracting 49 from itself leaves 0.

$$x^{2}-20x=-2-49$$

Subtract 49 from -2.

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

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.

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

Square -10.

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

Add -51 to 100.

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

Factor x^{2}-20x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-10=7 x-10=-7$$

Add 10 to both sides of the equation.

$$x=17 x=3$$