Question

solve by using the quadratic formula. round to the nearest tenth if necessary
X square -2x-15=0

Answer 1

\[x^2-2x-15=0\]
\[x_{1,2}={-b \pm \sqrt{b^2-4ac} \over 2a}\]

Answer 2

\[\LARGE x^2-2x-15=0\]
the quadratic formula is:
\[\LARGE x_{1/2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
in your case you have:

\[\LARGE \underbrace{1}_{a}\cdot x^2\;\;\underbrace{-2}_{b}x\;\;\underbrace{-15}_{c}=0\]
now substitute :)

Answer 3

wait, i dont get it

Answer 4

\[\LARGE x_{1/2}=\frac{-(-2)\pm\sqrt{(-2)^2-4\cdot 1\cdot (-15)}}{2\cdot 1}\] can you do it now? :)

Answer 5

not really, i know 2 square is gonna be 4 then multiply -4 times -15 which is 60, but then what do i do

Answer 6

Do you know what a square root is?

Answer 7

\[\LARGE x_{1/2}=\frac{2\pm\sqrt{4+60}}{2}\]
\[\LARGE x_{1/2}=\frac{2\pm\sqrt{64}}{2}\]
\[\LARGE x_{1/2}=\frac{2\pm\sqrt{8^2}}{2}\]
\[\LARGE x_{1/2}=\frac{2\pm8}{2}\]



\[\LARGE x_{1}=\frac{2-8}{2}=\frac{-6}{2}=-3\]


\[\LARGE x_{2}=\frac{2+8}{2}\]
do it :)

Answer 8

i dont get the two last ones , like how did you get 2-8 over 2 =-6 over 2 =-3