Solve 14x^2-9x-65=0

$$14 { x }^{ 2 } -9x-65 = 0$$

Solve for x

$$x = -\frac{13}{7} = -1\frac{6}{7} \approx -1.857142857$$

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

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 14x^{2}+ax+bx-65. To find a and b, set up a system to be solved.

$$a+b=-9 ab=14\left(-65\right)=-910$$

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 -910.

$$1,-910 2,-455 5,-182 7,-130 10,-91 13,-70 14,-65 26,-35$$

Calculate the sum for each pair.

$$1-910=-909 2-455=-453 5-182=-177 7-130=-123 10-91=-81 13-70=-57 14-65=-51 26-35=-9$$

The solution is the pair that gives sum -9.

$$a=-35 b=26$$

Rewrite 14x^{2}-9x-65 as \left(14x^{2}-35x\right)+\left(26x-65\right).

$$\left(14x^{2}-35x\right)+\left(26x-65\right)$$

Factor out 7x in the first and 13 in the second group.

$$7x\left(2x-5\right)+13\left(2x-5\right)$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(7x+13\right)$$

To find equation solutions, solve 2x-5=0 and 7x+13=0.

$$x=\frac{5}{2} x=-\frac{13}{7}$$

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.

$$14x^{2}-9x-65=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 14 for a, -9 for b, and -65 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 14\left(-65\right)}}{2\times 14}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 14\left(-65\right)}}{2\times 14}$$

Multiply -4 times 14.

$$x=\frac{-\left(-9\right)±\sqrt{81-56\left(-65\right)}}{2\times 14}$$

Multiply -56 times -65.

$$x=\frac{-\left(-9\right)±\sqrt{81+3640}}{2\times 14}$$

Add 81 to 3640.

$$x=\frac{-\left(-9\right)±\sqrt{3721}}{2\times 14}$$

Take the square root of 3721.

$$x=\frac{-\left(-9\right)±61}{2\times 14}$$

The opposite of -9 is 9.

$$x=\frac{9±61}{2\times 14}$$

Multiply 2 times 14.

$$x=\frac{9±61}{28}$$

Now solve the equation x=\frac{9±61}{28} when ± is plus. Add 9 to 61.

$$x=\frac{70}{28}$$

Reduce the fraction \frac{70}{28} to lowest terms by extracting and canceling out 14.

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

Now solve the equation x=\frac{9±61}{28} when ± is minus. Subtract 61 from 9.

$$x=-\frac{52}{28}$$

Reduce the fraction \frac{-52}{28} to lowest terms by extracting and canceling out 4.

$$x=-\frac{13}{7}$$

The equation is now solved.

$$x=\frac{5}{2} x=-\frac{13}{7}$$

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.

$$14x^{2}-9x-65=0$$

Add 65 to both sides of the equation.

$$14x^{2}-9x-65-\left(-65\right)=-\left(-65\right)$$

Subtracting -65 from itself leaves 0.

$$14x^{2}-9x=-\left(-65\right)$$

Subtract -65 from 0.

$$14x^{2}-9x=65$$

Divide both sides by 14.

$$\frac{14x^{2}-9x}{14}=\frac{65}{14}$$

Dividing by 14 undoes the multiplication by 14.

$$x^{2}-\frac{9}{14}x=\frac{65}{14}$$

Divide -\frac{9}{14}, the coefficient of the x term, by 2 to get -\frac{9}{28}. Then add the square of -\frac{9}{28} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{9}{14}x+\left(-\frac{9}{28}\right)^{2}=\frac{65}{14}+\left(-\frac{9}{28}\right)^{2}$$

Square -\frac{9}{28} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{14}x+\frac{81}{784}=\frac{65}{14}+\frac{81}{784}$$

Add \frac{65}{14} to \frac{81}{784} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{9}{14}x+\frac{81}{784}=\frac{3721}{784}$$

Factor x^{2}-\frac{9}{14}x+\frac{81}{784}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{28}\right)^{2}=\frac{3721}{784}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{28}\right)^{2}}=\sqrt{\frac{3721}{784}}$$

Simplify.

$$x-\frac{9}{28}=\frac{61}{28} x-\frac{9}{28}=-\frac{61}{28}$$

Add \frac{9}{28} to both sides of the equation.

$$x=\frac{5}{2} x=-\frac{13}{7}$$