Solve 2k^2+36=18k

$$2 k ^ { 2 } + 36 = 18 k$$

Solve for k

$$k=3$$

$$k=6$$

Steps Using Factoring By Grouping

Subtract 18k from both sides.

$$2k^{2}+36-18k=0$$

Divide both sides by 2.

$$k^{2}+18-9k=0$$

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

$$k^{2}-9k+18=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as k^{2}+ak+bk+18. To find a and b, set up a system to be solved.

$$a+b=-9 ab=1\times 18=18$$

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 18.

$$-1,-18 -2,-9 -3,-6$$

Calculate the sum for each pair.

$$-1-18=-19 -2-9=-11 -3-6=-9$$

The solution is the pair that gives sum -9.

$$a=-6 b=-3$$

Rewrite k^{2}-9k+18 as \left(k^{2}-6k\right)+\left(-3k+18\right).

$$\left(k^{2}-6k\right)+\left(-3k+18\right)$$

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

$$k\left(k-6\right)-3\left(k-6\right)$$

Factor out common term k-6 by using distributive property.

$$\left(k-6\right)\left(k-3\right)$$

To find equation solutions, solve k-6=0 and k-3=0.

$$k=6 k=3$$

Steps Using the Quadratic Formula

Subtract 18k from both sides.

$$2k^{2}+36-18k=0$$

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.

$$2k^{2}-18k+36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -18 for b, and 36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 2\times 36}}{2\times 2}$$

Square -18.

$$k=\frac{-\left(-18\right)±\sqrt{324-4\times 2\times 36}}{2\times 2}$$

Multiply -4 times 2.

$$k=\frac{-\left(-18\right)±\sqrt{324-8\times 36}}{2\times 2}$$

Multiply -8 times 36.

$$k=\frac{-\left(-18\right)±\sqrt{324-288}}{2\times 2}$$

Add 324 to -288.

$$k=\frac{-\left(-18\right)±\sqrt{36}}{2\times 2}$$

Take the square root of 36.

$$k=\frac{-\left(-18\right)±6}{2\times 2}$$

The opposite of -18 is 18.

$$k=\frac{18±6}{2\times 2}$$

Multiply 2 times 2.

$$k=\frac{18±6}{4}$$

Now solve the equation k=\frac{18±6}{4} when ± is plus. Add 18 to 6.

$$k=\frac{24}{4}$$

Divide 24 by 4.

$$k=6$$

Now solve the equation k=\frac{18±6}{4} when ± is minus. Subtract 6 from 18.

$$k=\frac{12}{4}$$

Divide 12 by 4.

$$k=3$$

The equation is now solved.

$$k=6 k=3$$

Steps for Completing the Square

Subtract 18k from both sides.

$$2k^{2}+36-18k=0$$

Subtract 36 from both sides. Anything subtracted from zero gives its negation.

$$2k^{2}-18k=-36$$

Divide both sides by 2.

$$\frac{2k^{2}-18k}{2}=-\frac{36}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$k^{2}+\left(-\frac{18}{2}\right)k=-\frac{36}{2}$$

Divide -18 by 2.

$$k^{2}-9k=-\frac{36}{2}$$

Divide -36 by 2.

$$k^{2}-9k=-18$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-9k+\left(-\frac{9}{2}\right)^{2}=-18+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-9k+\frac{81}{4}=-18+\frac{81}{4}$$

Add -18 to \frac{81}{4}.

$$k^{2}-9k+\frac{81}{4}=\frac{9}{4}$$

Factor k^{2}-9k+\frac{81}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-\frac{9}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{9}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$k-\frac{9}{2}=\frac{3}{2} k-\frac{9}{2}=-\frac{3}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$k=6 k=3$$