Solve 12j^2-5j-3=0

Question

Solve the quadratic equation

Solve by factoring

Solve using the quadratic formula

Solve by completing the square

j_{1} = - \frac{1}{3}, j_{2} = \frac{3}{4}

Alternative Form

j_{1} = - 0. \dot{3}, j_{2} = 0.75

Evaluate

12 j^{2} - 5 j - 3 = 0

Factor the expression

More Steps

Evaluate

12 j^{2} - 5 j - 3

Rewrite the expression

12 j^{2} + (- 9 + 4) j - 3

Calculate

12 j^{2} - 9 j + 4 j - 3

Rewrite the expression

3 j \times 4 j - 3 j \times 3 + 4 j - 3

Factor out 3 j from the expression

3 j (4 j - 3) + 4 j - 3

Factor out 4 j - 3 from the expression

(3 j + 1) (4 j - 3)

(3 j + 1) (4 j - 3) = 0

When the product of factors equals 0,at least one factor is 0

4 j - 3 = 0 3 j + 1 = 0

Solve the equation for j

More Steps

Evaluate

4 j - 3 = 0

Move the constant to the right-hand side and change its sign

4 j = 0 + 3

Removing 0 doesn't change the value,so remove it from the expression

4 j = 3

Divide both sides

\frac{4 j}{4} = \frac{3}{4}

Divide the numbers

j = \frac{3}{4}

j = \frac{3}{4} 3 j + 1 = 0

Solve the equation for j

More Steps

Evaluate

3 j + 1 = 0

Move the constant to the right-hand side and change its sign

3 j = 0 - 1

Removing 0 doesn't change the value,so remove it from the expression

3 j = - 1

Divide both sides

\frac{3 j}{3} = \frac{- 1}{3}

Divide the numbers

j = \frac{- 1}{3}

Use \frac{- a}{b} = \frac{a}{- b} = - \frac{a}{b} to rewrite the fraction

j = - \frac{1}{3}

j = \frac{3}{4} j = - \frac{1}{3}

Solution

j_{1} = - \frac{1}{3}, j_{2} = \frac{3}{4}

Alternative Form

j_{1} = - 0. \dot{3}, j_{2} = 0.75

Show Solution