Solve 6p^2+5=17p

$$6 p ^ { 2 } + 5 = 17 p$$

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 17p from both sides.

$$6p^{2}+5-17p=0$$

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

$$6p^{2}-17p+5=0$$

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.

$$a+b=-17 ab=6\times 5=30$$

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 30.

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

Calculate the sum for each pair.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

The solution is the pair that gives sum -17.

$$a=-15 b=-2$$

Rewrite 6p^{2}-17p+5 as \left(6p^{2}-15p\right)+\left(-2p+5\right).

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

Factor out 3p in the first and -1 in the second group.

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

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 17p from both sides.

$$6p^{2}+5-17p=0$$

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.

$$6p^{2}-17p+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -17 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 6\times 5}}{2\times 6}$$

Square -17.

$$p=\frac{-\left(-17\right)±\sqrt{289-4\times 6\times 5}}{2\times 6}$$

Multiply -4 times 6.

$$p=\frac{-\left(-17\right)±\sqrt{289-24\times 5}}{2\times 6}$$

Multiply -24 times 5.

$$p=\frac{-\left(-17\right)±\sqrt{289-120}}{2\times 6}$$

Add 289 to -120.

$$p=\frac{-\left(-17\right)±\sqrt{169}}{2\times 6}$$

Take the square root of 169.

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

The opposite of -17 is 17.

$$p=\frac{17±13}{2\times 6}$$

Multiply 2 times 6.

$$p=\frac{17±13}{12}$$

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

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

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{17±13}{12} when ± is minus. Subtract 13 from 17.

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

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 17p from both sides.

$$6p^{2}+5-17p=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$6p^{2}-17p=-5$$

Divide both sides by 6.

$$\frac{6p^{2}-17p}{6}=-\frac{5}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$p^{2}-\frac{17}{6}p=-\frac{5}{6}$$

Divide -\frac{17}{6}, the coefficient of the x term, by 2 to get -\frac{17}{12}. Then add the square of -\frac{17}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}-\frac{17}{6}p+\left(-\frac{17}{12}\right)^{2}=-\frac{5}{6}+\left(-\frac{17}{12}\right)^{2}$$

Square -\frac{17}{12} by squaring both the numerator and the denominator of the fraction.

$$p^{2}-\frac{17}{6}p+\frac{289}{144}=-\frac{5}{6}+\frac{289}{144}$$

Add -\frac{5}{6} to \frac{289}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$p^{2}-\frac{17}{6}p+\frac{289}{144}=\frac{169}{144}$$

Factor p^{2}-\frac{17}{6}p+\frac{289}{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}.

$$\left(p-\frac{17}{12}\right)^{2}=\frac{169}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p-\frac{17}{12}\right)^{2}}=\sqrt{\frac{169}{144}}$$

Simplify.

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

Add \frac{17}{12} to both sides of the equation.

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