Solve Using the Quadratic Formula x(5x+14)=3

$x (5 x + 14) = 3$

Step 1

Move all terms to the left side of the equation and simplify.

Simplify the left side.

Simplify $x (5 x + 14)$ .

Simplify by multiplying through.

Apply the distributive property.

$x (5 x) + x \cdot 14 = 3$

Reorder.

Rewrite using the commutative property of multiplication.

$5 x \cdot x + x \cdot 14 = 3$

Move $14$ to the left of $x$ .

$5 x \cdot x + 14 \cdot x = 3$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$5 (x \cdot x) + 14 \cdot x = 3$

Multiply $x$ by $x$ .

$5 x^{2} + 14 \cdot x = 3$

$5 x^{2} + 14 x = 3$

Subtract $3$ from both sides of the equation.

$5 x^{2} + 14 x - 3 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 5$ , $b = 14$ , and $c = - 3$ into the quadratic formula and solve for $x$ .

$\frac{- 14 \pm \sqrt{14^{2} - 4 \cdot (5 \cdot - 3)}}{2 \cdot 5}$

Step 4

Simplify.

Simplify the numerator.

Raise $14$ to the power of $2$ .

$x = \frac{- 14 \pm \sqrt{196 - 4 \cdot 5 \cdot - 3}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot - 3$ .

Multiply $- 4$ by $5$ .

$x = \frac{- 14 \pm \sqrt{196 - 20 \cdot - 3}}{2 \cdot 5}$

Multiply $- 20$ by $- 3$ .

$x = \frac{- 14 \pm \sqrt{196 + 60}}{2 \cdot 5}$

Add $196$ and $60$ .

$x = \frac{- 14 \pm \sqrt{256}}{2 \cdot 5}$

Rewrite $256$ as $16^{2}$ .

$x = \frac{- 14 \pm \sqrt{16^{2}}}{2 \cdot 5}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 14 \pm 16}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{- 14 \pm 16}{10}$

Simplify $\frac{- 14 \pm 16}{10}$ .

$x = \frac{- 7 \pm 8}{5}$

Step 5

The final answer is the combination of both solutions.

$x = \frac{1}{5}, - 3$

Solve Using the Quadratic Formula x(5x+14)=3

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