Write in Standard Form (2x-3)^2=-5

${(2 x - 3)}^{2} = - 5$

Step 1

Add $5$ to both sides of the equation.

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

Step 2

Simplify ${(2 x - 3)}^{2} + 5$ .

Simplify each term.

Rewrite ${(2 x - 3)}^{2}$ as $(2 x - 3) (2 x - 3)$ .

$(2 x - 3) (2 x - 3) + 5 = 0$

Expand $(2 x - 3) (2 x - 3)$ using the FOIL Method.

Apply the distributive property.

$2 x (2 x - 3) - 3 (2 x - 3) + 5 = 0$

Apply the distributive property.

$2 x (2 x) + 2 x \cdot - 3 - 3 (2 x - 3) + 5 = 0$

Apply the distributive property.

$2 x (2 x) + 2 x \cdot - 3 - 3 (2 x) - 3 \cdot - 3 + 5 = 0$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$2 \cdot 2 x \cdot x + 2 x \cdot - 3 - 3 (2 x) - 3 \cdot - 3 + 5 = 0$

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

Move $x$ .

$2 \cdot 2 (x \cdot x) + 2 x \cdot - 3 - 3 (2 x) - 3 \cdot - 3 + 5 = 0$

Multiply $x$ by $x$ .

$2 \cdot 2 x^{2} + 2 x \cdot - 3 - 3 (2 x) - 3 \cdot - 3 + 5 = 0$

Multiply $2$ by $2$ .

$4 x^{2} + 2 x \cdot - 3 - 3 (2 x) - 3 \cdot - 3 + 5 = 0$

Multiply $- 3$ by $2$ .

$4 x^{2} - 6 x - 3 (2 x) - 3 \cdot - 3 + 5 = 0$

Multiply $2$ by $- 3$ .

$4 x^{2} - 6 x - 6 x - 3 \cdot - 3 + 5 = 0$

Multiply $- 3$ by $- 3$ .

$4 x^{2} - 6 x - 6 x + 9 + 5 = 0$

Subtract $6 x$ from $- 6 x$ .

$4 x^{2} - 12 x + 9 + 5 = 0$

Add $9$ and $5$ .

$4 x^{2} - 12 x + 14 = 0$

Step 3

To write a polynomial in standard form, simplify and then arrange the terms in descending order.

$a x^{2} + b x + c = 0$

Step 4

The standard form is $4 x^{2} - 12 x + 14$ .

$4 x^{2} - 12 x + 14 = 0$

Step 5