Solve Using the Quadratic Formula x(5x+4)=1

$x (5 x + 4) = 1$

Step 1

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

Simplify the left side.

Simplify $x (5 x + 4)$ .

Simplify by multiplying through.

Apply the distributive property.

$x (5 x) + x \cdot 4 = 1$

Reorder.

Rewrite using the commutative property of multiplication.

$5 x \cdot x + x \cdot 4 = 1$

Move $4$ to the left of $x$ .

$5 x \cdot x + 4 \cdot x = 1$

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

Move $x$ .

$5 (x \cdot x) + 4 \cdot x = 1$

Multiply $x$ by $x$ .

$5 x^{2} + 4 \cdot x = 1$

$5 x^{2} + 4 x = 1$

Subtract $1$ from both sides of the equation.

$5 x^{2} + 4 x - 1 = 0$

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 $a = 5$ , $b = 4$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (5 \cdot - 1)}}{2 \cdot 5}$

Step 4

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 5 \cdot - 1}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot - 1$ .

Multiply $- 4$ by $5$ .

$x = \frac{- 4 \pm \sqrt{16 - 20 \cdot - 1}}{2 \cdot 5}$

Multiply $- 20$ by $- 1$ .

$x = \frac{- 4 \pm \sqrt{16 + 20}}{2 \cdot 5}$

Add $16$ and $20$ .

$x = \frac{- 4 \pm \sqrt{36}}{2 \cdot 5}$

Rewrite $36$ as $6^{2}$ .

$x = \frac{- 4 \pm \sqrt{6^{2}}}{2 \cdot 5}$

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

$x = \frac{- 4 \pm 6}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{- 4 \pm 6}{10}$

Simplify $\frac{- 4 \pm 6}{10}$ .

$x = \frac{- 2 \pm 3}{5}$

Step 5

The final answer is the combination of both solutions.

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

Solve Using the Quadratic Formula x(5x+4)=1

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