Solve 2x^2=13x+45

$$2 x ^ { 2 } = 13 x + 45$$

Solve for x

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

$$x=9$$

Steps Using Factoring By Grouping

Subtract 13x from both sides.

$$2x^{2}-13x=45$$

Subtract 45 from both sides.

$$2x^{2}-13x-45=0$$

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

$$a+b=-13 ab=2\left(-45\right)=-90$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -90.

$$1,-90 2,-45 3,-30 5,-18 6,-15 9,-10$$

Calculate the sum for each pair.

$$1-90=-89 2-45=-43 3-30=-27 5-18=-13 6-15=-9 9-10=-1$$

The solution is the pair that gives sum -13.

$$a=-18 b=5$$

Rewrite 2x^{2}-13x-45 as \left(2x^{2}-18x\right)+\left(5x-45\right).

$$\left(2x^{2}-18x\right)+\left(5x-45\right)$$

Factor out 2x in the first and 5 in the second group.

$$2x\left(x-9\right)+5\left(x-9\right)$$

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

$$\left(x-9\right)\left(2x+5\right)$$

To find equation solutions, solve x-9=0 and 2x+5=0.

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

Steps Using the Quadratic Formula

Subtract 13x from both sides.

$$2x^{2}-13x=45$$

Subtract 45 from both sides.

$$2x^{2}-13x-45=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -13 for b, and -45 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 2\left(-45\right)}}{2\times 2}$$

Square -13.

$$x=\frac{-\left(-13\right)±\sqrt{169-4\times 2\left(-45\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-13\right)±\sqrt{169-8\left(-45\right)}}{2\times 2}$$

Multiply -8 times -45.

$$x=\frac{-\left(-13\right)±\sqrt{169+360}}{2\times 2}$$

Add 169 to 360.

$$x=\frac{-\left(-13\right)±\sqrt{529}}{2\times 2}$$

Take the square root of 529.

$$x=\frac{-\left(-13\right)±23}{2\times 2}$$

The opposite of -13 is 13.

$$x=\frac{13±23}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{13±23}{4}$$

Now solve the equation x=\frac{13±23}{4} when ± is plus. Add 13 to 23.

$$x=\frac{36}{4}$$

Divide 36 by 4.

$$x=9$$

Now solve the equation x=\frac{13±23}{4} when ± is minus. Subtract 23 from 13.

$$x=-\frac{10}{4}$$

Reduce the fraction \frac{-10}{4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 13x from both sides.

$$2x^{2}-13x=45$$

Divide both sides by 2.

$$\frac{2x^{2}-13x}{2}=\frac{45}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{13}{2}x=\frac{45}{2}$$

Divide -\frac{13}{2}, the coefficient of the x term, by 2 to get -\frac{13}{4}. Then add the square of -\frac{13}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{13}{2}x+\left(-\frac{13}{4}\right)^{2}=\frac{45}{2}+\left(-\frac{13}{4}\right)^{2}$$

Square -\frac{13}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{13}{2}x+\frac{169}{16}=\frac{45}{2}+\frac{169}{16}$$

Add \frac{45}{2} to \frac{169}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{13}{2}x+\frac{169}{16}=\frac{529}{16}$$

Factor x^{2}-\frac{13}{2}x+\frac{169}{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}.

$$\left(x-\frac{13}{4}\right)^{2}=\frac{529}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{4}\right)^{2}}=\sqrt{\frac{529}{16}}$$

Simplify.

$$x-\frac{13}{4}=\frac{23}{4} x-\frac{13}{4}=-\frac{23}{4}$$

Add \frac{13}{4} to both sides of the equation.

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