Solve 19p+6p^2=-3

$$19 p + 6 p ^ { 2 } = - 3$$

Solve for p

$$p=-3$$

$$p=-\frac{1}{6}\approx -0.166666667$$

Steps Using Factoring By Grouping

Add 3 to both sides.

$$19p+6p^{2}+3=0$$

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

$$6p^{2}+19p+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6p^{2}+ap+bp+3. To find a and b, set up a system to be solved.

$$a+b=19 ab=6\times 3=18$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 18.

$$1,18 2,9 3,6$$

Calculate the sum for each pair.

$$1+18=19 2+9=11 3+6=9$$

The solution is the pair that gives sum 19.

$$a=1 b=18$$

Rewrite 6p^{2}+19p+3 as \left(6p^{2}+p\right)+\left(18p+3\right).

$$\left(6p^{2}+p\right)+\left(18p+3\right)$$

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

$$p\left(6p+1\right)+3\left(6p+1\right)$$

Factor out common term 6p+1 by using distributive property.

$$\left(6p+1\right)\left(p+3\right)$$

To find equation solutions, solve 6p+1=0 and p+3=0.

$$p=-\frac{1}{6} p=-3$$

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.

$$6p^{2}+19p=-3$$

Add 3 to both sides of the equation.

$$6p^{2}+19p-\left(-3\right)=-3-\left(-3\right)$$

Subtracting -3 from itself leaves 0.

$$6p^{2}+19p-\left(-3\right)=0$$

Subtract -3 from 0.

$$6p^{2}+19p+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 19 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-19±\sqrt{19^{2}-4\times 6\times 3}}{2\times 6}$$

Square 19.

$$p=\frac{-19±\sqrt{361-4\times 6\times 3}}{2\times 6}$$

Multiply -4 times 6.

$$p=\frac{-19±\sqrt{361-24\times 3}}{2\times 6}$$

Multiply -24 times 3.

$$p=\frac{-19±\sqrt{361-72}}{2\times 6}$$

Add 361 to -72.

$$p=\frac{-19±\sqrt{289}}{2\times 6}$$

Take the square root of 289.

$$p=\frac{-19±17}{2\times 6}$$

Multiply 2 times 6.

$$p=\frac{-19±17}{12}$$

Now solve the equation p=\frac{-19±17}{12} when ± is plus. Add -19 to 17.

$$p=-\frac{2}{12}$$

Reduce the fraction \frac{-2}{12} to lowest terms by extracting and canceling out 2.

$$p=-\frac{1}{6}$$

Now solve the equation p=\frac{-19±17}{12} when ± is minus. Subtract 17 from -19.

$$p=-\frac{36}{12}$$

Divide -36 by 12.

$$p=-3$$

The equation is now solved.

$$p=-\frac{1}{6} p=-3$$

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.

$$6p^{2}+19p=-3$$

Divide both sides by 6.

$$\frac{6p^{2}+19p}{6}=-\frac{3}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$p^{2}+\frac{19}{6}p=-\frac{3}{6}$$

Reduce the fraction \frac{-3}{6} to lowest terms by extracting and canceling out 3.

$$p^{2}+\frac{19}{6}p=-\frac{1}{2}$$

Divide \frac{19}{6}, the coefficient of the x term, by 2 to get \frac{19}{12}. Then add the square of \frac{19}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}+\frac{19}{6}p+\left(\frac{19}{12}\right)^{2}=-\frac{1}{2}+\left(\frac{19}{12}\right)^{2}$$

Square \frac{19}{12} by squaring both the numerator and the denominator of the fraction.

$$p^{2}+\frac{19}{6}p+\frac{361}{144}=-\frac{1}{2}+\frac{361}{144}$$

Add -\frac{1}{2} to \frac{361}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$p^{2}+\frac{19}{6}p+\frac{361}{144}=\frac{289}{144}$$

Factor p^{2}+\frac{19}{6}p+\frac{361}{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}.

$$\left(p+\frac{19}{12}\right)^{2}=\frac{289}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p+\frac{19}{12}\right)^{2}}=\sqrt{\frac{289}{144}}$$

Simplify.

$$p+\frac{19}{12}=\frac{17}{12} p+\frac{19}{12}=-\frac{17}{12}$$

Subtract \frac{19}{12} from both sides of the equation.

$$p=-\frac{1}{6} p=-3$$