Question 162005
{{{2x^2-3x+1=0}}}

We know that the last term must be 1

Set up the equation like this
it could be -1,-1 or 1,1

Because you have a negative middle term it is most likely -1,-1

(2x-1)(x-1)=0

Therefore the solution is x= 1/2 or 1 we need to make the equation=0 and the only way to accomplish this is to make one or both terms =0




{{{4x^2+9x=9}}}

We need to subtract 9 from both sides of the equation 

4x^2+9x-9=0

We need to find two numbers that have a product of 9 the possibilities are 3,3
9,1
We have a negative number so we know that the product is -

We then need to know what 2 numbers give a product of 4

2,2 or 4,1

Let's set up the equation

using 3,3

(_x-3)(_x+3)
The middle terms need to add up to 9x Let's try 4 and 1

(4x-3)(x+3)  The middle term for this problem is -3x+12x=9x

We have the correct factors
the solution is x=3/4 or -3



{{{2y^2+y=-11y}}}

Here we add 11y to both sides of the equation

2y^2+y+11y=0

2y^2+12y=0

factor out 2
2(y^2+6)=0
y=sqrt6 i
remember that i=sqrt of -1


{{{y^2-7-44=0}}}

First we try to calculate the numbers that generate a product of 44

4,11 or 44,1

Let's try 4,11

(y-11)(y+4)

y =11 or -4


6x^2+11x-1=0
ax^2+bx+c=0 a=6 b=11 c=-1*[invoke quadratic "x", 6, 11, -1]




Use the quadratic formula