Final answer:
The solutions to 3x²+3x+5=0 are imaginary. The zeros of y=x²-4x-5 are x=5 and x=-1.
Step-by-step explanation:
To find the solution to the equation 3x²+3x+5=0, we can use the quadratic formula.
The quadratic formula is x = (-b ± sqrt(b² - 4ac))/2a, where a, b, and c are the coefficients of the quadratic equation.
For the equation 3x²+3x+5=0, the coefficients are a=3, b=3, and c=5. Substituting these values into the quadratic formula, we get x = (-3 ± sqrt(3² - 4*3*5))/(2*3).
Simplifying the equation gives us two solutions: x = (-3 ± sqrt(-39))/6.
Since the discriminant (b² - 4ac) is negative, there are no real solutions to the equation. Therefore, the solution to 3x²+3x+5=0 is imaginary.
For the equation y=x²-4x-5=0, we can find the zeros by factoring or using the quadratic formula.
By factoring, we re-write the equation as (x-5)(x+1)=0. Setting each factor equal to zero, we get x-5=0 and x+1=0. Solving these equations, we find x=5 and x=-1.