Question

3x^2+2=-8x <--- solve by the quadratic formula , set the equation to "0" first. must be in simplest radical form, show work please

Answer 1

coefficients are
a = 3
b = 8
c = 2

substitute these into quadratic formula

Answer 2

how i get the equation to equal to zero first?

Answer 3

just add 8x to both sides.

Answer 4

i got x=\[\pm8\]\[\sqrt{8^2-4(3)(2)}\] /2(3)
is that right?

Answer 5

Should be \[x = \frac{-8 \pm \sqrt{8^2-4(3)(2)}}{2(3)}\]

So yeah, the 8 out front is negative, and the plus or minus goes between the -b and the square root. Otherwise that's right.

Answer 6

so what would i do next?

Answer 7

Do the multiplication in the square root part and the denominator, then simplify it.

Answer 8

x=-8 \[\pm\] \[\sqrt{8^2-36}\]/6

Answer 9

is that right?

Answer 10

4(3)(2) = 24\[\implies \sqrt{8^2 - 4(3)(2)} \]\[= \sqrt{8^2 - 24} \]\[= \sqrt{64 - 24}\]\[=\sqrt{40}\]\[=\sqrt{4\cdot 10} \]\[=2\sqrt{10}\]

Answer 11

so you can replace the square root you have with 2 root(10)

Answer 12

how you get 24?

Answer 13

would we put the -8 and that upside down tee looking like before the square root problems you did?

Answer 14

that upside-down t is called the plus or minus sign. and yes.