Solve for j 3j^2-20j=-12

$3 j^{2} - 20 j = - 12$

Add $12$ to both sides of the equation.

$3 j^{2} - 20 j + 12 = 0$

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 $a \cdot c = 3 \cdot 12 = 36$ and whose sum is $b = - 20$ .

Factor $- 20$ out of $- 20 j$ .

$3 j^{2} - 20 j + 12 = 0$

Rewrite $- 20$ as $- 2$ plus $- 18$

$3 j^{2} + (- 2 - 18) j + 12 = 0$

Apply the distributive property.

$3 j^{2} - 2 j - 18 j + 12 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 j^{2} - 2 j) - 18 j + 12 = 0$

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

$j (3 j - 2) - 6 (3 j - 2) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 j - 2$ .

$(3 j - 2) (j - 6) = 0$

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

$3 j - 2 = 0$

$j - 6 = 0$

Set $3 j - 2$ equal to $0$ and solve for $j$ .

Set $3 j - 2$ equal to $0$ .

$3 j - 2 = 0$

Solve $3 j - 2 = 0$ for $j$ .

Add $2$ to both sides of the equation.

$3 j = 2$

Divide each term in $3 j = 2$ by $3$ and simplify.

Divide each term in $3 j = 2$ by $3$ .

$\frac{3 j}{3} = \frac{2}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 j}{3} = \frac{2}{3}$

Divide $j$ by $1$ .

$j = \frac{2}{3}$

Set $j - 6$ equal to $0$ and solve for $j$ .

Set $j - 6$ equal to $0$ .

$j - 6 = 0$

Add $6$ to both sides of the equation.

$j = 6$

The final solution is all the values that make $(3 j - 2) (j - 6) = 0$ true.

$j = \frac{2}{3}, 6$

The result can be shown in multiple forms.

Exact Form:

$j = \frac{2}{3}, 6$

Decimal Form:

$j = 0.\overline{6}, 6$

Solve for j 3j^2-20j=-12

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