Question

How do you find the roots, real and imaginary, of `y=-5(x-2)^2+3x^2-10x-5` using the quadratic formula?

Answer 1

`x=5/2 +- 5/2i`

Explanation:

`y=−5(x−2)^2+3x^2−10x−5`

by expanding and simplifing,
`y=−5(x^2−4x+4)+3x^2−10x−5`
`y=−5x^2+20x-20+3x^2−10x−5`
`y=−2x^2+10x-25`

Using the quadratic formula: `x=(-b+- sqrt(b^2-4ac))/(2a)`,
`x=(-10+- sqrt(10^2-4(-2)(-25)))/(2(-2))`

Simplifying,
`x=5/2 +- 5/2i`

Answer 2

We first work this into a 'proper' quadratic equation:

Explanation:

`y=-5(x^2-4x+4)+3x^2-10x-5->`

`y=-5x^2+20x-20+3x^2-10x-5->`

`y=-2x^2+10x-25`

Now we may use the formula:

`x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)`

`x_(1,2)=(-10+-sqrt(10^2-4*(-2) * (-25)))/(2*(-2))`

`x_(1,2)=(-10+-sqrt(100-200))/(-4)=(-10+-sqrt(-100))/(-4)=(-10+-10i)/(-4)`

`x_(1,2)=(10+-10i)/4=2 1/2+-2 1/2ior2 1/2(1+-i)`

You can see from the graph that there are no real roots.
graph{-2x^2+10x-25 [-85.1, 102.3, -78, 15.76]}