Question

How do you find the roots, real and imaginary, of `y=-2x^2 + 7x +12(x/2-1)^2` using the quadratic formula?

Answer 1

`x_1=5/2+sqrt23/2i~~2.5+2.4i`
`x_2=5/2-sqrt23/2i~~2.5-2.4i`

When `5/2` in real and `sqrt23/2` is imaginary for both

Explanation:

for the form of:
`y=ax^2+bx+c`

the quadratic formula is:
`x_(1,2)=(-b+-sqrt{b^2-4*a*c})/(2*a)`

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`y=-2x^2+7x+12(x/2-1)^2`

`=> y=-2x^2+7x+12(x^2/4+2*(x/2)*(-1)+1)`

`=> y=-2x^2+7x+12(x^2/4-x+1)`

`=> y=-2x^2+7x+3x^2-12x+12`

`=> y=x^2-5x+12`

`=>`
`a=1`
`b=-5`
`c=12`

`=>`
`x_(1,2)=(-b+-sqrt{b^2-4*a*c})/(2*a)=(-(-5)+-sqrt{25-4*1*12})/(2*1)=`
`=(5+-sqrt{25-48})/(2)=`
`=(5+-sqrt{-23})/(2)=`
`=(5+-sqrt23i)/(2)=`
`=>`
`x_1=(5+sqrt23i)/(2)=5/2+sqrt23/2i=2.5+sqrt23/2i~~2.5+2.4i`
`x_2=(5-sqrt23i)/(2)=5/2-sqrt23/2i=2.5-sqrt23/2i~~2.5-2.4i`