Find the Roots (Zeros) 7x^2+16x-48

$7 x^{2} + 16 x - 48$

Step 1

Set $7 x^{2} + 16 x - 48$ equal to $0$ .

$7 x^{2} + 16 x - 48 = 0$

Step 2

Solve for $x$ .

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 $a \cdot c = 7 \cdot - 48 = - 336$ and whose sum is $b = 16$ .

Factor $16$ out of $16 x$ .

$7 x^{2} + 16 (x) - 48 = 0$

Rewrite $16$ as $- 12$ plus $28$

$7 x^{2} + (- 12 + 28) x - 48 = 0$

Apply the distributive property.

$7 x^{2} - 12 x + 28 x - 48 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(7 x^{2} - 12 x) + 28 x - 48 = 0$

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

$x (7 x - 12) + 4 (7 x - 12) = 0$

Factor the polynomial by factoring out the greatest common factor, $7 x - 12$ .

$(7 x - 12) (x + 4) = 0$

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

$7 x - 12 = 0$

$x + 4 = 0$

Set $7 x - 12$ equal to $0$ and solve for $x$ .

Set $7 x - 12$ equal to $0$ .

$7 x - 12 = 0$

Solve $7 x - 12 = 0$ for $x$ .

Add $12$ to both sides of the equation.

$7 x = 12$

Divide each term in $7 x = 12$ by $7$ and simplify.

Divide each term in $7 x = 12$ by $7$ .

$\frac{7 x}{7} = \frac{12}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{12}{7}$

Divide $x$ by $1$ .

$x = \frac{12}{7}$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $(7 x - 12) (x + 4) = 0$ true.

$x = \frac{12}{7}, - 4$

Step 3

Find the Roots (Zeros) 7x^2+16x-48

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