How do you find the roots, real and imaginary, of $y = - 5 x^{2} + 40 x - 34$ $y=-5x^2 +40x -34$ using the quadratic formula?

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Answer 1 · 2016-01-26T20:51:37

$4 \pm \sqrt{9.2}$ $4+-sqrt(9.2)$

$\frac{- b \pm \sqrt{b^{2} - 4 \cdot a \cdot c}}{2 \cdot a}$ $(-b+-sqrt(b^2-4*a*c))/(2*a)$

with a=-5, b=40 and c=-34 for this particular equation

$\frac{- 40 \pm \sqrt{40^{2} - 4 \cdot (- 5) (- 34)}}{2 \cdot (- 5)}$ $(-40+-sqrt(40^2-4*(-5)(-34)))/(2*(-5))$ , which gives:

$\frac{- 40 \pm \sqrt{1600 - 680}}{- 10}$ $(-40+-sqrt(1600-680))/(-10)$ ,

$\frac{- 40 \pm \sqrt{920}}{- 10}$ $(-40+-sqrt(920))/(-10)$ ,

$\frac{40 \pm \sqrt{920}}{10}$ $(40+-sqrt(920))/(10)$ ,

As 920 is not a perfect square, you can symplify the expression in several ways

$\frac{40 \pm \sqrt{4 \cdot 230}}{10} = \frac{20 \pm \sqrt{230}}{5} = 4 \pm \sqrt{9.2}$ $(40+-sqrt(4*230))/(10)=(20+-sqrt(230))/(5)=4+-sqrt(9.2)$

How do you find the roots, real and imaginary, of y=-5x^2 +40x -34 y=−5x2+40x−34y=-5x^2 +40x -34 using the quadratic formula?