Solve 6p^2-5=13p

$$6 p ^ { 2 } - 5 = 13 p$$

Solve for p

$$p=-\frac{1}{3}\approx -0.333333333$$

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

Steps Using Factoring By Grouping

Subtract 13p from both sides.

$$6p^{2}-5-13p=0$$

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

$$6p^{2}-13p-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6p^{2}+ap+bp-5. To find a and b, set up a system to be solved.

$$a+b=-13 ab=6\left(-5\right)=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -13.

$$a=-15 b=2$$

Rewrite 6p^{2}-13p-5 as \left(6p^{2}-15p\right)+\left(2p-5\right).

$$\left(6p^{2}-15p\right)+\left(2p-5\right)$$

Factor out 3p in 6p^{2}-15p.

$$3p\left(2p-5\right)+2p-5$$

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

$$\left(2p-5\right)\left(3p+1\right)$$

To find equation solutions, solve 2p-5=0 and 3p+1=0.

$$p=\frac{5}{2} p=-\frac{1}{3}$$

Steps Using the Quadratic Formula

Subtract 13p from both sides.

$$6p^{2}-5-13p=0$$

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.

$$6p^{2}-13p-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -13 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 6\left(-5\right)}}{2\times 6}$$

Square -13.

$$p=\frac{-\left(-13\right)±\sqrt{169-4\times 6\left(-5\right)}}{2\times 6}$$

Multiply -4 times 6.

$$p=\frac{-\left(-13\right)±\sqrt{169-24\left(-5\right)}}{2\times 6}$$

Multiply -24 times -5.

$$p=\frac{-\left(-13\right)±\sqrt{169+120}}{2\times 6}$$

Add 169 to 120.

$$p=\frac{-\left(-13\right)±\sqrt{289}}{2\times 6}$$

Take the square root of 289.

$$p=\frac{-\left(-13\right)±17}{2\times 6}$$

The opposite of -13 is 13.

$$p=\frac{13±17}{2\times 6}$$

Multiply 2 times 6.

$$p=\frac{13±17}{12}$$

Now solve the equation p=\frac{13±17}{12} when ± is plus. Add 13 to 17.

$$p=\frac{30}{12}$$

Reduce the fraction \frac{30}{12} to lowest terms by extracting and canceling out 6.

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

Now solve the equation p=\frac{13±17}{12} when ± is minus. Subtract 17 from 13.

$$p=-\frac{4}{12}$$

Reduce the fraction \frac{-4}{12} to lowest terms by extracting and canceling out 4.

$$p=-\frac{1}{3}$$

The equation is now solved.

$$p=\frac{5}{2} p=-\frac{1}{3}$$

Steps for Completing the Square

Subtract 13p from both sides.

$$6p^{2}-5-13p=0$$

Add 5 to both sides. Anything plus zero gives itself.

$$6p^{2}-13p=5$$

Divide both sides by 6.

$$\frac{6p^{2}-13p}{6}=\frac{5}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$p^{2}-\frac{13}{6}p=\frac{5}{6}$$

Divide -\frac{13}{6}, the coefficient of the x term, by 2 to get -\frac{13}{12}. Then add the square of -\frac{13}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}-\frac{13}{6}p+\left(-\frac{13}{12}\right)^{2}=\frac{5}{6}+\left(-\frac{13}{12}\right)^{2}$$

Square -\frac{13}{12} by squaring both the numerator and the denominator of the fraction.

$$p^{2}-\frac{13}{6}p+\frac{169}{144}=\frac{5}{6}+\frac{169}{144}$$

Add \frac{5}{6} to \frac{169}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$p^{2}-\frac{13}{6}p+\frac{169}{144}=\frac{289}{144}$$

Factor p^{2}-\frac{13}{6}p+\frac{169}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(p-\frac{13}{12}\right)^{2}=\frac{289}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p-\frac{13}{12}\right)^{2}}=\sqrt{\frac{289}{144}}$$

Simplify.

$$p-\frac{13}{12}=\frac{17}{12} p-\frac{13}{12}=-\frac{17}{12}$$

Add \frac{13}{12} to both sides of the equation.

$$p=\frac{5}{2} p=-\frac{1}{3}$$