Question

2(g^2+8g+16)=0

Answer 1

Distribute the 2 first

Answer 2

I did

Answer 3

Wait y
The original equation is 2g^2+16g+32=0

Answer 4

oh, are you solving for g, right?

Answer 5

Yea

Answer 6

since its all equal to 0 you can use the quadratic formula, do you know that?

Answer 7

Uh...

Answer 8

\(\large\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) is that equation familiar?

Answer 9

Yes it is but I dunno if it would help

Answer 10

Its the best way to solve for it imo, or you could graph

Answer 11

Ok...

Answer 12

I'll show you both ways just in case

\(\large\frac{-b\pm\sqrt{b^2-4ac}}{2a}\); a=2, b=16, c=32

\(\large\frac{-(16)+\sqrt{(16)^2-4\cdot2\cdot32}}{2\cdot2}\)=\(\large\frac{-16+\sqrt{256-256}}{4}\)= \(\large\frac{-16+\color{red}{\sqrt{0}}}{4}\)=\(\frac{-16}{4}\)=-4
and
\(\large\frac{-(16)-\sqrt{(16)^2-4\cdot2\cdot32}}{2\cdot2}\)=\(\large\frac{-16-\sqrt{256-256}}{4}\)=\(\large\frac{-16+\color{red}{\sqrt{0}}}{4}\)=\(\frac{-16}{4}\)=-4

Answer 13

Graph

Answer 14

It shouldn’t be a parabola

Answer 15

\(2g^2+16g+32=0\) is in parabolic form o.O
\(ax^2+bx+c\) = parabolic form

Answer 16

I’m supposed to mind what numbers are x to equal 0

Answer 17

@dude well done my guy.

And @silvernight269 quadratic formulas are for quadratic equations, which tend to form a parabola when they are graphed. Quadratic equations always have this base form, where a, b, and c are all numbers:

\(ax^2 + bx+c=0\)

Answer 18

just factor my guy

Answer 19

Oh ur right. Forgot🤦🏻♂️

Answer 20

g^2+8+16 is just (g+4)^2

Answer 21

8g*

Answer 22

@silvernight269 let us know if you have more questions about the process :) dude did an excellent job guiding you here.

Answer 23

Ok