Solve 2x^2-18x+81-53=0

$$2 x ^ { 2 } - 18 x + 81 - 53 = 0$$

Solve for x

$$x=2$$

$$x=7$$

Steps Using Factoring By Grouping

Subtract 53 from 81 to get 28.

$$2x^{2}-18x+28=0$$

Divide both sides by 2.

$$x^{2}-9x+14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+14. To find a and b, set up a system to be solved.

$$a+b=-9 ab=1\times 14=14$$

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 14.

$$-1,-14 -2,-7$$

Calculate the sum for each pair.

$$-1-14=-15 -2-7=-9$$

The solution is the pair that gives sum -9.

$$a=-7 b=-2$$

Rewrite x^{2}-9x+14 as \left(x^{2}-7x\right)+\left(-2x+14\right).

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

Factor out x in the first and -2 in the second group.

$$x\left(x-7\right)-2\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(x-2\right)$$

To find equation solutions, solve x-7=0 and x-2=0.

$$x=7 x=2$$

Steps Using the Quadratic Formula

Subtract 53 from 81 to get 28.

$$2x^{2}-18x+28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -18 for b, and 28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 2\times 28}}{2\times 2}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 2\times 28}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-18\right)±\sqrt{324-8\times 28}}{2\times 2}$$

Multiply -8 times 28.

$$x=\frac{-\left(-18\right)±\sqrt{324-224}}{2\times 2}$$

Add 324 to -224.

$$x=\frac{-\left(-18\right)±\sqrt{100}}{2\times 2}$$

Take the square root of 100.

$$x=\frac{-\left(-18\right)±10}{2\times 2}$$

The opposite of -18 is 18.

$$x=\frac{18±10}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{18±10}{4}$$

Now solve the equation x=\frac{18±10}{4} when ± is plus. Add 18 to 10.

$$x=\frac{28}{4}$$

Divide 28 by 4.

$$x=7$$

Now solve the equation x=\frac{18±10}{4} when ± is minus. Subtract 10 from 18.

$$x=\frac{8}{4}$$

Divide 8 by 4.

$$x=2$$

The equation is now solved.

$$x=7 x=2$$

Steps for Completing the Square

Subtract 53 from 81 to get 28.

$$2x^{2}-18x+28=0$$

Subtract 28 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}-18x=-28$$

Divide both sides by 2.

$$\frac{2x^{2}-18x}{2}=-\frac{28}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{18}{2}\right)x=-\frac{28}{2}$$

Divide -18 by 2.

$$x^{2}-9x=-\frac{28}{2}$$

Divide -28 by 2.

$$x^{2}-9x=-14$$

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.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=-14+\left(-\frac{9}{2}\right)^{2}$$

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

$$x^{2}-9x+\frac{81}{4}=-14+\frac{81}{4}$$

Add -14 to \frac{81}{4}.

$$x^{2}-9x+\frac{81}{4}=\frac{25}{4}$$

Factor x^{2}-9x+\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}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{5}{2} x-\frac{9}{2}=-\frac{5}{2}$$

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

$$x=7 x=2$$