jeffry is wrong because it only works for 4
let's say jef was right
we would get (4)(4)=4, then 16=4 which is false, go back to school, jeff
we will solve to find out if george is right later (he's wrong because he set it 0 when it is equal to 4)
anwyay, the way to do it is expand, make one side 0 and then factor, then set each factor to 0
so
expanding we get
2x²+5x-3=4
minus 4 both sides
2x²+5x-7=0
ok, we need to use the ac method or something
for ax²+bx+c=0 where a is not eual to 0, multiply a and c and call the result r, find what 2 numbers multiply to get r and add to get b in the equation
2x²+5x-7=0
2 times -7=-14
what 2 numbers multiply to get 5 and add to get -14
-2 and 7
split middle number up
2x²-2x+7x-7=0
group
(2x²-2x)+(7x-7)=0
factor
2x(x-1)+7(x-1)=0
undistribute
(2x+7)(x-1)=0
set each to 0
2x+7=0
2x=-7
x=-7/2
x-1=0
x=1
x=-7/2 and 1
that's how it works
to check
x=-7/2
(2(-7/2)-1)((-7/2)+3)=4
(-7-1)((-7/2)+(6/2))=4
(-8)(-1/2)=4
8/2=4
4=4
true
x=1
(2(1)-1)(1+3)=4
(2-1)(4)=4
(1)(4)=4
4=4
true
x=-7/2 and 1
george and jeff are wrong
