y=5x^2+6x-2y=6x^2-3x+4y=9x^2+18x+9what is zero

To find the zeros of each equation, we need to set y equal to zero and solve for x.

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.

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.
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