Solve (6k-5)(2k-3)

Question

Simplify the expression

12 k^{2} - 28 k + 15

Evaluate

(6 k - 5) (2 k - 3)

Apply the distributive property

6 k \times 2 k - 6 k \times 3 - 5 \times 2 k - (- 5 \times 3)

Multiply the terms

More Steps

Evaluate

6 k \times 2 k

Multiply the numbers

12 k \times k

Multiply the terms

12 k^{2}

12 k^{2} - 6 k \times 3 - 5 \times 2 k - (- 5 \times 3)

Multiply the numbers

12 k^{2} - 18 k - 5 \times 2 k - (- 5 \times 3)

Multiply the numbers

12 k^{2} - 18 k - 10 k - (- 5 \times 3)

Multiply the numbers

12 k^{2} - 18 k - 10 k - (- 15)

If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses

12 k^{2} - 18 k - 10 k + 15

Solution

More Steps

Evaluate

- 18 k - 10 k

Collect like terms by calculating the sum or difference of their coefficients

(- 18 - 10) k

Subtract the numbers

- 28 k

12 k^{2} - 28 k + 15

Show Solution

Find the roots

k_{1} = \frac{5}{6}, k_{2} = \frac{3}{2}

Alternative Form

k_{1} = 0.8 \dot{3}, k_{2} = 1.5

Evaluate

(6 k - 5) (2 k - 3)

To find the roots of the expression,set the expression equal to 0

(6 k - 5) (2 k - 3) = 0

Separate the equation into 2 possible cases

6 k - 5 = 0 2 k - 3 = 0

Solve the equation

More Steps

Evaluate

6 k - 5 = 0

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

6 k = 0 + 5

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

6 k = 5

Divide both sides

\frac{6 k}{6} = \frac{5}{6}

Divide the numbers

k = \frac{5}{6}

k = \frac{5}{6} 2 k - 3 = 0

Solve the equation

More Steps

Evaluate

2 k - 3 = 0

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

2 k = 0 + 3

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

2 k = 3

Divide both sides

\frac{2 k}{2} = \frac{3}{2}

Divide the numbers

k = \frac{3}{2}

k = \frac{5}{6} k = \frac{3}{2}

Solution

k_{1} = \frac{5}{6}, k_{2} = \frac{3}{2}

Alternative Form

k_{1} = 0.8 \dot{3}, k_{2} = 1.5

Show Solution