Solve for r 15r^2-6r-3=-2r

$15 r^{2} - 6 r - 3 = - 2 r$

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

Add $2 r$ to both sides of the equation.

$15 r^{2} - 6 r - 3 + 2 r = 0$

Add $- 6 r$ and $2 r$ .

$15 r^{2} - 4 r - 3 = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 15 \cdot - 3 = - 45$ and whose sum is $b = - 4$ .

Factor $- 4$ out of $- 4 r$ .

$15 r^{2} - 4 r - 3 = 0$

Rewrite $- 4$ as $5$ plus $- 9$

$15 r^{2} + (5 - 9) r - 3 = 0$

Apply the distributive property.

$15 r^{2} + 5 r - 9 r - 3 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(15 r^{2} + 5 r) - 9 r - 3 = 0$

Factor out the greatest common factor (GCF) from each group.

$5 r (3 r + 1) - 3 (3 r + 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 r + 1$ .

$(3 r + 1) (5 r - 3) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$3 r + 1 = 0$

$5 r - 3 = 0$

Set $3 r + 1$ equal to $0$ and solve for $r$ .

Set $3 r + 1$ equal to $0$ .

$3 r + 1 = 0$

Solve $3 r + 1 = 0$ for $r$ .

Subtract $1$ from both sides of the equation.

$3 r = - 1$

Divide each term in $3 r = - 1$ by $3$ and simplify.

Divide each term in $3 r = - 1$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $r$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Set $5 r - 3$ equal to $0$ and solve for $r$ .

Set $5 r - 3$ equal to $0$ .

$5 r - 3 = 0$

Solve $5 r - 3 = 0$ for $r$ .

Add $3$ to both sides of the equation.

$5 r = 3$

Divide each term in $5 r = 3$ by $5$ and simplify.

Divide each term in $5 r = 3$ by $5$ .

$\frac{5 r}{5} = \frac{3}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 r}{5} = \frac{3}{5}$

Divide $r$ by $1$ .

$r = \frac{3}{5}$

The final solution is all the values that make $(3 r + 1) (5 r - 3) = 0$ true.

$r = - \frac{1}{3}, \frac{3}{5}$

The result can be shown in multiple forms.

Exact Form:

$r = - \frac{1}{3}, \frac{3}{5}$

Decimal Form:

$r = - 0.\overline{3}, 0.6$

Solve for r 15r^2-6r-3=-2r

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