Solve x^2+(-9)^2-8x+10*(-9)+25=0

$$x ^ { 2 } + ( - 9 ) ^ { 2 } - 8 x + 10 \cdot ( - 9 ) + 25 = 0$$

Solve for x

$$x=4$$

Steps Using Factoring

Calculate -9 to the power of 2 and get 81.

$$x^{2}+81-8x+10\left(-9\right)+25=0$$

Multiply 10 and -9 to get -90.

$$x^{2}+81-8x-90+25=0$$

Subtract 90 from 81 to get -9.

$$x^{2}-9-8x+25=0$$

Add -9 and 25 to get 16.

$$x^{2}+16-8x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}-8x+16=0$$

To solve the equation, factor x^{2}-8x+16 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.

$$a+b=-8 ab=16$$

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 16.

$$-1,-16 -2,-8 -4,-4$$

Calculate the sum for each pair.

$$-1-16=-17 -2-8=-10 -4-4=-8$$

The solution is the pair that gives sum -8.

$$a=-4 b=-4$$

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

$$\left(x-4\right)\left(x-4\right)$$

Rewrite as a binomial square.

$$\left(x-4\right)^{2}$$

To find equation solution, solve x-4=0.

$$x=4$$

Steps Using Factoring By Grouping

Calculate -9 to the power of 2 and get 81.

$$x^{2}+81-8x+10\left(-9\right)+25=0$$

Multiply 10 and -9 to get -90.

$$x^{2}+81-8x-90+25=0$$

Subtract 90 from 81 to get -9.

$$x^{2}-9-8x+25=0$$

Add -9 and 25 to get 16.

$$x^{2}+16-8x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}-8x+16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+16. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\times 16=16$$

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 16.

$$-1,-16 -2,-8 -4,-4$$

Calculate the sum for each pair.

$$-1-16=-17 -2-8=-10 -4-4=-8$$

The solution is the pair that gives sum -8.

$$a=-4 b=-4$$

Rewrite x^{2}-8x+16 as \left(x^{2}-4x\right)+\left(-4x+16\right).

$$\left(x^{2}-4x\right)+\left(-4x+16\right)$$

Factor out x in the first and -4 in the second group.

$$x\left(x-4\right)-4\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x-4\right)$$

Rewrite as a binomial square.

$$\left(x-4\right)^{2}$$

To find equation solution, solve x-4=0.

$$x=4$$

Steps Using the Quadratic Formula

Calculate -9 to the power of 2 and get 81.

$$x^{2}+81-8x+10\left(-9\right)+25=0$$

Multiply 10 and -9 to get -90.

$$x^{2}+81-8x-90+25=0$$

Subtract 90 from 81 to get -9.

$$x^{2}-9-8x+25=0$$

Add -9 and 25 to get 16.

$$x^{2}+16-8x=0$$

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.

$$x^{2}-8x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 16}}{2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 16}}{2}$$

Multiply -4 times 16.

$$x=\frac{-\left(-8\right)±\sqrt{64-64}}{2}$$

Add 64 to -64.

$$x=\frac{-\left(-8\right)±\sqrt{0}}{2}$$

Take the square root of 0.

$$x=-\frac{-8}{2}$$

The opposite of -8 is 8.

$$x=\frac{8}{2}$$

Divide 8 by 2.

$$x=4$$

Steps for Completing the Square

Calculate -9 to the power of 2 and get 81.

$$x^{2}+81-8x+10\left(-9\right)+25=0$$

Multiply 10 and -9 to get -90.

$$x^{2}+81-8x-90+25=0$$

Subtract 90 from 81 to get -9.

$$x^{2}-9-8x+25=0$$

Add -9 and 25 to get 16.

$$x^{2}+16-8x=0$$

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-8x=-16$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=-16+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-16+16$$

Add -16 to 16.

$$x^{2}-8x+16=0$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-4\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-4=0 x-4=0$$

Add 4 to both sides of the equation.

$$x=4 x=4$$

The equation is now solved. Solutions are the same.

$$x=4$$