-2x^2+6x-3=0 with the quadratic formula

someone plz help

a = -2  b = 6  c = -3

x = -6 +- sq root (6^2 -4 * -2 *-3) / 2 * -2

x1 = [-6 + sq root (36 - 24)] / -4

x1 = [-6 + sq root (36 - 24)] / -4

x1 = (-6 + sq root (12) ) / -4

x1 = (-6 + 3.4641016151 ) / -4

x1 = -2.5358983849  / -4

x1 = 0.6339745962

x2 = (-6 - sq root (12) ) / -4

x2 = (-6 - sq root (12) ) / -4

x2= (-6 - 3.4641016151 ) / -4

x2 = - 9.4641016151 / -4

x2 = 2.3660254038

Step-by-step explanation:

no this is a nonlinear negative

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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