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the quadratic formula is -b (±)Sqrt[b-4ac]/2a\[-b \pm \frac{ \sqrt{b-4ac}}{ 2a}\]

where
ax^2 +bx +c
just substitute and you will get your answer.

I have all that work! I just get confused with the last part? like would i subtract 36-(-160)

36--160?
well double negative is positive,
so?
36+160

36-(-160) is the same as 36+160

so it would become 196?

it would

|dw:1352933217966:dw|

so would that be the answer?

you can keep going and there's a +- between the 6 and square root

you should have
\[\Large x = \frac{6\pm\sqrt{196}}{8}\]
but you can keep going

(6*14)/8
=(3*14)/4
=(3*7)/2
=21/2

yes i have that!

okay, what is wrong?

wait where did you get the 6*14/8

your most welcome,