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

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Answer 1 · 2017-04-28T10:16:47

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

$y=−5(x−2)^2+3x^2−10x−5$

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 · 2017-04-28T10:22:08

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

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

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

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

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]}

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