Question

solve
(x-2)(x+1)=5
step by step please

Answer 1

\[(x+2)(x+1)=5\]
\[x^2+x+2x+2-5=0\]
\[x^2+3x-3=0\]
\[D=3\cdot3-4\cdot(-3)\cdot1=21\]
\[x_{1,2}=\frac{-3\pm\sqrt{21}}{2}\]

Answer 2

Anyhow, this is how you do a product...
(a+b)(c+d) = a*c+b*d+a*d+b*c.

Answer 3

x * x = x^2
x * 1 = 1x
-2 * x = -2x
-2 * 1 = -2
Then add them all up and simplify where necessary: x^2-x-7 = 0
Then: x^2-x = 7
And from here I'm stuck aslo :L

Answer 4

thomas i think there was suppose to be a minus between x and 2
\[(x-2)(x+1)=5\]
\[x^2-2x+x-2=5\]
\[x^2-x-2-5=0\]
\[x^2-x-7=0\]
\[x=\frac{1 \pm \sqrt{1-4(1)(-7)}}{2}\]

Answer 5

@myininaya oh yeah sorry, i missed it ><