How do you find the zeros, real and imaginary, of $y=3x^2+17x-43$ using the quadratic formula?

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Answer 1 · 2016-07-18T06:42:48

We have real zeros $(-17-sqrt805)/6$ or $(-17+sqrt805)/6$

Zeros of $y=3x^2+17x-43$ are those values of $x$ for which $y=0$ . Hence, zeros are given by the solution of quadratic equation $3x^2+17x-43=0$ .

Quadratic formula gives solution of equation $ax^2+bx+c=0$ as $x=(-b+-sqrt(b^2-4ac))/(2a)$ . Hence for $3x^2+17x-43=0$ , we have

$x=(-17+-sqrt(17^2-4*3*(-43)))/(2*3)$

= $(-17+-sqrt(289+516))/6$

= $(-17+-sqrt805)/6$

= $(-17-sqrt805)/6$ or $(-17+sqrt805)/6$ ,

Which are real zeros of $3x^2+17x-43=0$ .

How do you find the zeros, real and imaginary, of y=3x^2+17x-43y=3x^2+17x-43 using the quadratic formula?