Solve by Factoring x^2-5x+33=7x-3

$x^{2} - 5 x + 33 = 7 x - 3$

Step 1

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

Subtract $7 x$ from both sides of the equation.

$x^{2} - 5 x + 33 - 7 x = - 3$

Add $3$ to both sides of the equation.

$x^{2} - 5 x + 33 - 7 x + 3 = 0$

Step 2

Simplify $x^{2} - 5 x + 33 - 7 x + 3$ .

Subtract $7 x$ from $- 5 x$ .

$x^{2} - 12 x + 33 + 3 = 0$

Add $33$ and $3$ .

$x^{2} - 12 x + 36 = 0$

Step 3

Factor using the perfect square rule.

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

$x^{2} - 12 x + 6^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$12 x = 2 \cdot x \cdot 6$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 6$ .

${(x - 6)}^{2} = 0$

Step 4

Set the $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 5

Add $6$ to both sides of the equation.

$x = 6$