Find the Zeros of Quadratic Equations: Step-by-Step Guide

To find the zeros of each equation, we need to set y equal to zero and solve for x.
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y = 5x^2 + 6x - 2
Setting y = 0:
0 = 5x^2 + 6x - 2
Using the quadratic formula: x = (-b ± √(b^2 - 4ac))/(2a)
x = (-6 ± √(6^2 - 4(5)(-2)))/(2(5))
x = (-6 ± √(36 + 40))/10
x = (-6 ± √(76))/10
x = (-6 ± √(4*19))/10
x = (-6 ± 2√19)/10
x = (-3 ± √19)/5

Therefore, the zeros of the first equation are x = (-3 + √19)/5 and x = (-3 - √19)/5.
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y = 6x^2 - 3x + 4
Setting y = 0:
0 = 6x^2 - 3x + 4
This equation does not have any real solutions, so there are no zeros for this equation.
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y = 9x^2 + 18x + 9
Setting y = 0:
0 = 9x^2 + 18x + 9
Dividing by 9:
0 = x^2 + 2x + 1
This equation can be factored as:
0 = (x + 1)^2
Setting each factor equal to zero:
x + 1 = 0
x = -1

Therefore, the zero of the third equation is x = -1.