Find the Intersection of the Functions f(x)=x^2-2x+3 ; f(x)=-2x+28

$f (x) = x^{2} - 2 x + 3$ ; $f (x) = - 2 x + 28$

Substitute $- 2 x + 28$ for $f (x)$ .

$- 2 x + 28 = x^{2} - 2 x + 3$

Solve for $x$ .

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x^{2} - 2 x + 3 = - 2 x + 28$

Move all terms containing $x$ to the left side of the equation.

Add $2 x$ to both sides of the equation.

$x^{2} - 2 x + 3 + 2 x = 28$

Combine the opposite terms in $x^{2} - 2 x + 3 + 2 x$ .

Add $- 2 x$ and $2 x$ .

$x^{2} + 0 + 3 = 28$

Add $x^{2}$ and $0$ .

$x^{2} + 3 = 28$

Move all terms not containing $x$ to the right side of the equation.

Subtract $3$ from both sides of the equation.

$x^{2} = 28 - 3$

Subtract $3$ from $28$ .

$x^{2} = 25$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{25}$

Simplify $\pm \sqrt{25}$ .

Rewrite $25$ as $5^{2}$ .

$x = \pm \sqrt{5^{2}}$

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

$x = \pm 5$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 5$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 5$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 5, - 5$