Question

NEED HELP RN PLZ!!!!!!!!!!!!!!!!!!!!!Complete the square to find the zeros of y=x^2+12x-5?

Answer 1

Answer: -6 +/- $\sqrt{41}$ = x

Step-by-step explanation:

   y = x² + 12x - 5

 +5                +5

y + 5 = x² + 12x              

+ 36                +36           added (12 ÷ 2)² to both sides

y + 41 = (x + 6)²

To find the zeros, let y = 0:

0 + 41 = (x + 6)²

$\sqrt{41} = (\sqrt{x+6} )^{2}$

$+/-\sqrt{41}$ = x + 6

-6                   -6  

-6 +/- $\sqrt{41}$ = x