Solve Using the Quadratic Formula 10x-1=(2x+1)^2

$10 x - 1 = {(2 x + 1)}^{2}$

Step 1

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

Simplify the right side.

Simplify ${(2 x + 1)}^{2}$ .

Rewrite ${(2 x + 1)}^{2}$ as $(2 x + 1) (2 x + 1)$ .

$10 x - 1 = (2 x + 1) (2 x + 1)$

Expand $(2 x + 1) (2 x + 1)$ using the FOIL Method.

Apply the distributive property.

$10 x - 1 = 2 x (2 x + 1) + 1 (2 x + 1)$

Apply the distributive property.

$10 x - 1 = 2 x (2 x) + 2 x \cdot 1 + 1 (2 x + 1)$

Apply the distributive property.

$10 x - 1 = 2 x (2 x) + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$10 x - 1 = 2 \cdot 2 x \cdot x + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1$

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

Move $x$ .

$10 x - 1 = 2 \cdot 2 (x \cdot x) + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1$

Multiply $x$ by $x$ .

$10 x - 1 = 2 \cdot 2 x^{2} + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1$

Multiply $2$ by $2$ .

$10 x - 1 = 4 x^{2} + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1$

Multiply $2$ by $1$ .

$10 x - 1 = 4 x^{2} + 2 x + 1 (2 x) + 1 \cdot 1$

Multiply $2 x$ by $1$ .

$10 x - 1 = 4 x^{2} + 2 x + 2 x + 1 \cdot 1$

Multiply $1$ by $1$ .

$10 x - 1 = 4 x^{2} + 2 x + 2 x + 1$

Add $2 x$ and $2 x$ .

$10 x - 1 = 4 x^{2} + 4 x + 1$

Move all the expressions to the left side of the equation.

Subtract $4 x^{2}$ from both sides of the equation.

$10 x - 1 - 4 x^{2} = 4 x + 1$

Subtract $4 x$ from both sides of the equation.

$10 x - 1 - 4 x^{2} - 4 x = 1$

Subtract $1$ from both sides of the equation.

$10 x - 1 - 4 x^{2} - 4 x - 1 = 0$

Simplify $10 x - 1 - 4 x^{2} - 4 x - 1$ .

Subtract $4 x$ from $10 x$ .

$6 x - 1 - 4 x^{2} - 1 = 0$

Subtract $1$ from $- 1$ .

$6 x - 4 x^{2} - 2 = 0$

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 $a = - 4$ , $b = 6$ , and $c = - 2$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (- 4 \cdot - 2)}}{2 \cdot - 4}$

Step 4

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot - 4 \cdot - 2}}{2 \cdot - 4}$

Multiply $- 4 \cdot - 4 \cdot - 2$ .

Multiply $- 4$ by $- 4$ .

$x = \frac{- 6 \pm \sqrt{36 + 16 \cdot - 2}}{2 \cdot - 4}$

Multiply $16$ by $- 2$ .

$x = \frac{- 6 \pm \sqrt{36 - 32}}{2 \cdot - 4}$

Subtract $32$ from $36$ .

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

Rewrite $4$ as $2^{2}$ .

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

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

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

Multiply $2$ by $- 4$ .

$x = \frac{- 6 \pm 2}{- 8}$

Simplify $\frac{- 6 \pm 2}{- 8}$ .

$x = \frac{3 \pm 1}{4}$

Step 5

The final answer is the combination of both solutions.

$x = 1, \frac{1}{2}$

Solve Using the Quadratic Formula 10x-1=(2x+1)^2

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