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

##### 2 Answers
Apr 28, 2017

$x = \frac{5}{2} \pm \frac{5}{2} i$

#### 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: $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$,
$x = \frac{- 10 \pm \sqrt{{10}^{2} - 4 \left(- 2\right) \left(- 25\right)}}{2 \left(- 2\right)}$

Simplifying,
$x = \frac{5}{2} \pm \frac{5}{2} i$

Apr 28, 2017

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

#### Explanation:

$y = - 5 \left({x}^{2} - 4 x + 4\right) + 3 {x}^{2} - 10 x - 5 \to$

$y = - 5 {x}^{2} + 20 x - 20 + 3 {x}^{2} - 10 x - 5 \to$

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

Now we may use the formula:

${x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

${x}_{1 , 2} = \frac{- 10 \pm \sqrt{{10}^{2} - 4 \cdot \left(- 2\right) \cdot \left(- 25\right)}}{2 \cdot \left(- 2\right)}$

${x}_{1 , 2} = \frac{- 10 \pm \sqrt{100 - 200}}{- 4} = \frac{- 10 \pm \sqrt{- 100}}{- 4} = \frac{- 10 \pm 10 i}{- 4}$

${x}_{1 , 2} = \frac{10 \pm 10 i}{4} = 2 \frac{1}{2} \pm 2 \frac{1}{2} i \mathmr{and} 2 \frac{1}{2} \left(1 \pm i\right)$

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