Solve 0=x^2 +6x+13 by completing the square

No real solutions

Step-by-step explanation:

0=x^2+6x+13

Step 1: Subtract x^2 from both sides.

0−x^2=x^2+6x+13−x^2

−x^2=6x+13

Step 2: Subtract 6x from both sides.

−x^2−6x=6x+13−6x

−x^2−6x=13

Step 3: Since the coefficient of -x^2 is -1, divide both sides by -1.

−x^2−6x/−1 = 13/−1

x^2+6x=−13

Step 4: The coefficient of 6x is 6. Let b=6.

Then we need to add (b/2)^2=9 to both sides to complete the square.

Add 9 to both sides.

x^2+6x+9=−13+9

x^2+6x+9=−4

Step 5: Factor left side.

(x+3^)2=−4

Step 6: Take square root.

x+3=±√−4

Step 7: Add -3 to both sides.

x+3+−3=−3±√−4

x=−3±√−4

yes because there is exactly one y value for every x value

for function there can't be more than one value for every x value