How do you find the roots, real and imaginary, of $y=-(x-3)^2+4x^2-x-5$ using the quadratic formula?

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Answer 1 · 2016-04-04T13:15:25

Roots are real (irrational) numbers $(-5+sqrt193)/6$ and $(-5-sqrt193)/6$

Simplifying $y=-(x-3)^2+4x^2-x-5=-(x^2-6x+9)+4x^2-x-5$

or $y=3x^2+5x-14$

Its root will be given by the equation $3x^2+5x-14=0$ .

Using quadratic formula $(-b+-sqrt(b^2-4ac))/(2a)$

these are $(-5+-sqrt(5^2-4*3*(-14)))/(2*3)=(-5+-sqrt(25+168))/6=(-5+-sqrt193)/6$

Hence roots are irrational numbers $(-5+sqrt193)/6$ and $(-5-sqrt193)/6$

How do you find the roots, real and imaginary, of y=-(x-3)^2+4x^2-x-5 y=-(x-3)^2+4x^2-x-5 using the quadratic formula?