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

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Answer 1 · 2018-04-25T07:30:43

$x=1or4.5$

First, we expand $-(x-3)^2$ to get $-(x^2-6x+9)=-x^2+6x-9$

$y=-x^2+5x-x^2+6x-9=-2x^2+11x-9$

$x=(-11+-sqrt(11^2-4(-2*-9)))/(2*-2)$

$x=-(-11+-sqrt(121-4(18)))/4$

$x=-(-11+-sqrt(49))/4$

$x=-(-11+7)/4or-(-11-7)/4$

$x=4/4or18/4$

$x=1or4.5$

Graph:
graph{-x^2+5x-(x-3)^2 [-2.045, 7.955, -0.68, 4.32]}

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