Solve 2x^2-13x=-20

$$2 { x }^{ 2 } -13x=-20$$

Solve for x

$$x=4$$

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

Steps Using Factoring By Grouping

Add 20 to both sides.

$$2x^{2}-13x+20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+20. To find a and b, set up a system to be solved.

$$a+b=-13 ab=2\times 20=40$$

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 40.

$$-1,-40 -2,-20 -4,-10 -5,-8$$

Calculate the sum for each pair.

$$-1-40=-41 -2-20=-22 -4-10=-14 -5-8=-13$$

The solution is the pair that gives sum -13.

$$a=-8 b=-5$$

Rewrite 2x^{2}-13x+20 as \left(2x^{2}-8x\right)+\left(-5x+20\right).

$$\left(2x^{2}-8x\right)+\left(-5x+20\right)$$

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

$$2x\left(x-4\right)-5\left(x-4\right)$$

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

$$\left(x-4\right)\left(2x-5\right)$$

To find equation solutions, solve x-4=0 and 2x-5=0.

$$x=4 x=\frac{5}{2}$$

Steps Using the Quadratic Formula

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.

$$2x^{2}-13x=-20$$

Add 20 to both sides of the equation.

$$2x^{2}-13x-\left(-20\right)=-20-\left(-20\right)$$

Subtracting -20 from itself leaves 0.

$$2x^{2}-13x-\left(-20\right)=0$$

Subtract -20 from 0.

$$2x^{2}-13x+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -13 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 2\times 20}}{2\times 2}$$

Square -13.

$$x=\frac{-\left(-13\right)±\sqrt{169-4\times 2\times 20}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-13\right)±\sqrt{169-8\times 20}}{2\times 2}$$

Multiply -8 times 20.

$$x=\frac{-\left(-13\right)±\sqrt{169-160}}{2\times 2}$$

Add 169 to -160.

$$x=\frac{-\left(-13\right)±\sqrt{9}}{2\times 2}$$

Take the square root of 9.

$$x=\frac{-\left(-13\right)±3}{2\times 2}$$

The opposite of -13 is 13.

$$x=\frac{13±3}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{13±3}{4}$$

Now solve the equation x=\frac{13±3}{4} when ± is plus. Add 13 to 3.

$$x=\frac{16}{4}$$

Divide 16 by 4.

$$x=4$$

Now solve the equation x=\frac{13±3}{4} when ± is minus. Subtract 3 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.

$$x=4 x=\frac{5}{2}$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}-13x=-20$$

Divide both sides by 2.

$$\frac{2x^{2}-13x}{2}=-\frac{20}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{13}{2}x=-\frac{20}{2}$$

Divide -20 by 2.

$$x^{2}-\frac{13}{2}x=-10$$

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.

$$x^{2}-\frac{13}{2}x+\left(-\frac{13}{4}\right)^{2}=-10+\left(-\frac{13}{4}\right)^{2}$$

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

$$x^{2}-\frac{13}{2}x+\frac{169}{16}=-10+\frac{169}{16}$$

Add -10 to \frac{169}{16}.

$$x^{2}-\frac{13}{2}x+\frac{169}{16}=\frac{9}{16}$$

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}.

$$\left(x-\frac{13}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x-\frac{13}{4}=\frac{3}{4} x-\frac{13}{4}=-\frac{3}{4}$$

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

$$x=4 x=\frac{5}{2}$$