Solve x^2+45=14x

$$x ^ { 2 } + 45 = 14 x$$

Solve for x

$$x=5$$

$$x=9$$

Steps Using Factoring

Subtract 14x from both sides.

$$x^{2}+45-14x=0$$

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

$$x^{2}-14x+45=0$$

To solve the equation, factor x^{2}-14x+45 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.

$$a+b=-14 ab=45$$

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 45.

$$-1,-45 -3,-15 -5,-9$$

Calculate the sum for each pair.

$$-1-45=-46 -3-15=-18 -5-9=-14$$

The solution is the pair that gives sum -14.

$$a=-9 b=-5$$

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

$$\left(x-9\right)\left(x-5\right)$$

To find equation solutions, solve x-9=0 and x-5=0.

$$x=9 x=5$$

Steps Using Factoring By Grouping

Subtract 14x from both sides.

$$x^{2}+45-14x=0$$

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

$$x^{2}-14x+45=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+45. To find a and b, set up a system to be solved.

$$a+b=-14 ab=1\times 45=45$$

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 45.

$$-1,-45 -3,-15 -5,-9$$

Calculate the sum for each pair.

$$-1-45=-46 -3-15=-18 -5-9=-14$$

The solution is the pair that gives sum -14.

$$a=-9 b=-5$$

Rewrite x^{2}-14x+45 as \left(x^{2}-9x\right)+\left(-5x+45\right).

$$\left(x^{2}-9x\right)+\left(-5x+45\right)$$

Factor out x in the first and -5 in the second group.

$$x\left(x-9\right)-5\left(x-9\right)$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(x-5\right)$$

To find equation solutions, solve x-9=0 and x-5=0.

$$x=9 x=5$$

Steps Using the Quadratic Formula

Subtract 14x from both sides.

$$x^{2}+45-14x=0$$

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.

$$x^{2}-14x+45=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -14 for b, and 45 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 45}}{2}$$

Square -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\times 45}}{2}$$

Multiply -4 times 45.

$$x=\frac{-\left(-14\right)±\sqrt{196-180}}{2}$$

Add 196 to -180.

$$x=\frac{-\left(-14\right)±\sqrt{16}}{2}$$

Take the square root of 16.

$$x=\frac{-\left(-14\right)±4}{2}$$

The opposite of -14 is 14.

$$x=\frac{14±4}{2}$$

Now solve the equation x=\frac{14±4}{2} when ± is plus. Add 14 to 4.

$$x=\frac{18}{2}$$

Divide 18 by 2.

$$x=9$$

Now solve the equation x=\frac{14±4}{2} when ± is minus. Subtract 4 from 14.

$$x=\frac{10}{2}$$

Divide 10 by 2.

$$x=5$$

The equation is now solved.

$$x=9 x=5$$

Steps for Completing the Square

Subtract 14x from both sides.

$$x^{2}+45-14x=0$$

Subtract 45 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-14x=-45$$

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

$$x^{2}-14x+\left(-7\right)^{2}=-45+\left(-7\right)^{2}$$

Square -7.

$$x^{2}-14x+49=-45+49$$

Add -45 to 49.

$$x^{2}-14x+49=4$$

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

$$\left(x-7\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-7\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-7=2 x-7=-2$$

Add 7 to both sides of the equation.

$$x=9 x=5$$