What are the zeros of the following quadratic equation: y = 6x^2 - 17x - 3 step by step

  x = 3, x = -1/6 are the zeros

Step-by-step explanation:

I find graphing the equation using a graphing calculator to be about the fastest way to find the zeros.

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For this equation, you can look for factors of 6·(-3) = -18 that have a sum of -17. Those are -18 and +1, so the factorization of the equation is ...

  y = (1/6)(6x -18)(6x +1) = (x -3)(6x +1)

The roots are the values of x that make the factors be zero, so x=3 and x=-1/6.

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You can also use the quadratic formula to find the zeros. That tells you the solution to ...

  ax² +bx +c = 0

is

  x = (-b ±√(b²-4ac))/(2a)

Comparing your equation to the standard form, you can identify the coefficients as ...

  a = 6, b = -17, c = -3

so the zeros are ...

  x = (-(-17) ±√((-17)² -4(6)(-3)))/(2(6))

  x = (17 ±√369)/12 = (17 ±19)/12 = {-2, 36}/12 = {-1/6, 3}

The zeros are x = -1/6 and x = 3.

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step-by-step explanation: