Solve 3p^2-5p-28=0

Question

Solve the quadratic equation

Solve by factoring

Solve using the quadratic formula

Solve by completing the square

p_{1} = - \frac{7}{3}, p_{2} = 4

Alternative Form

p_{1} = - 2. \dot{3}, p_{2} = 4

Evaluate

3 p^{2} - 5 p - 28 = 0

Factor the expression

More Steps

Evaluate

3 p^{2} - 5 p - 28

Rewrite the expression

3 p^{2} + (7 - 12) p - 28

Calculate

3 p^{2} + 7 p - 12 p - 28

Rewrite the expression

p \times 3 p + p \times 7 - 4 \times 3 p - 4 \times 7

Factor out p from the expression

p (3 p + 7) - 4 \times 3 p - 4 \times 7

Factor out - 4 from the expression

p (3 p + 7) - 4 (3 p + 7)

Factor out 3 p + 7 from the expression

(p - 4) (3 p + 7)

(p - 4) (3 p + 7) = 0

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

p - 4 = 0 3 p + 7 = 0

Solve the equation for p

More Steps

Evaluate

p - 4 = 0

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

p = 0 + 4

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

p = 4

p = 4 3 p + 7 = 0

Solve the equation for p

More Steps

Evaluate

3 p + 7 = 0

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

3 p = 0 - 7

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

3 p = - 7

Divide both sides

\frac{3 p}{3} = \frac{- 7}{3}

Divide the numbers

p = \frac{- 7}{3}

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

p = - \frac{7}{3}

p = 4 p = - \frac{7}{3}

Solution

p_{1} = - \frac{7}{3}, p_{2} = 4

Alternative Form

p_{1} = - 2. \dot{3}, p_{2} = 4

Show Solution