Solve for p 31p^2+10p+1=6p^2

$31 p^{2} + 10 p + 1 = 6 p^{2}$

Move all terms containing $p$ to the left side of the equation.

Subtract $6 p^{2}$ from both sides of the equation.

$31 p^{2} + 10 p + 1 - 6 p^{2} = 0$

Subtract $6 p^{2}$ from $31 p^{2}$ .

$25 p^{2} + 10 p + 1 = 0$

Factor using the perfect square rule.

Rewrite $25 p^{2}$ as ${(5 p)}^{2}$ .

${(5 p)}^{2} + 10 p + 1 = 0$

Rewrite $1$ as $1^{2}$ .

${(5 p)}^{2} + 10 p + 1^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$10 p = 2 \cdot (5 p) \cdot 1$

Rewrite the polynomial.

${(5 p)}^{2} + 2 \cdot (5 p) \cdot 1 + 1^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 5 p$ and $b = 1$ .

${(5 p + 1)}^{2} = 0$

Set the $5 p + 1$ equal to $0$ .

$5 p + 1 = 0$

Solve for $p$ .

Subtract $1$ from both sides of the equation.

$5 p = - 1$

Divide each term in $5 p = - 1$ by $5$ and simplify.

Divide each term in $5 p = - 1$ by $5$ .

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

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

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

Divide $p$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$p = - 0.2$