Find the Roots (Zeros) (2y+1)^2-4y^2=5

${(2 y + 1)}^{2} - 4 y^{2} = 5$

Step 1

Solve for $y$ .

Subtract $5$ from both sides of the equation.

${(2 y + 1)}^{2} - 4 y^{2} - 5 = 0$

Simplify ${(2 y + 1)}^{2} - 4 y^{2} - 5$ .

Simplify each term.

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

$(2 y + 1) (2 y + 1) - 4 y^{2} - 5 = 0$

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

Apply the distributive property.

$2 y (2 y + 1) + 1 (2 y + 1) - 4 y^{2} - 5 = 0$

Apply the distributive property.

$2 y (2 y) + 2 y \cdot 1 + 1 (2 y + 1) - 4 y^{2} - 5 = 0$

Apply the distributive property.

$2 y (2 y) + 2 y \cdot 1 + 1 (2 y) + 1 \cdot 1 - 4 y^{2} - 5 = 0$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$2 \cdot 2 y \cdot y + 2 y \cdot 1 + 1 (2 y) + 1 \cdot 1 - 4 y^{2} - 5 = 0$

Multiply $y$ by $y$ by adding the exponents.

Move $y$ .

$2 \cdot 2 (y \cdot y) + 2 y \cdot 1 + 1 (2 y) + 1 \cdot 1 - 4 y^{2} - 5 = 0$

Multiply $y$ by $y$ .

$2 \cdot 2 y^{2} + 2 y \cdot 1 + 1 (2 y) + 1 \cdot 1 - 4 y^{2} - 5 = 0$

Multiply $2$ by $2$ .

$4 y^{2} + 2 y \cdot 1 + 1 (2 y) + 1 \cdot 1 - 4 y^{2} - 5 = 0$

Multiply $2$ by $1$ .

$4 y^{2} + 2 y + 1 (2 y) + 1 \cdot 1 - 4 y^{2} - 5 = 0$

Multiply $2 y$ by $1$ .

$4 y^{2} + 2 y + 2 y + 1 \cdot 1 - 4 y^{2} - 5 = 0$

Multiply $1$ by $1$ .

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

Add $2 y$ and $2 y$ .

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

Combine the opposite terms in $4 y^{2} + 4 y + 1 - 4 y^{2} - 5$ .

Subtract $4 y^{2}$ from $4 y^{2}$ .

$4 y + 1 + 0 - 5 = 0$

Add $4 y + 1$ and $0$ .

$4 y + 1 - 5 = 0$

Subtract $5$ from $1$ .

$4 y - 4 = 0$

Factor $4$ out of $4 y - 4$ .

Factor $4$ out of $4 y$ .

$4 (y) - 4 = 0$

Factor $4$ out of $- 4$ .

$4 (y) + 4 (- 1) = 0$

Factor $4$ out of $4 (y) + 4 (- 1)$ .

$4 (y - 1) = 0$

Divide each term in $4 (y - 1) = 0$ by $4$ and simplify.

Divide each term in $4 (y - 1) = 0$ by $4$ .

$\frac{4 (y - 1)}{4} = \frac{0}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 (y - 1)}{4} = \frac{0}{4}$

Divide $y - 1$ by $1$ .

$y - 1 = \frac{0}{4}$

Simplify the right side.

Divide $0$ by $4$ .

$y - 1 = 0$

Add $1$ to both sides of the equation.

$y = 1$

Step 2

Set $5$ equal to $0$ .

$1 = 0$

Step 3

Since $1 \neq 0$ , there are no solutions.

No solution

Find the Roots (Zeros) (2y+1)^2-4y^2=5

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