Final answer:
The quadratic equation 3y^2+5y-10=0 has two real solutions. These can be determined by using the quadratic formula, yielding two values for y = -2.84 or y= 1.173
Step-by-step explanation:
The solutions to the given quadratic equation 3y^2+5y-10=0 can be found using the quadratic formula, which is y = (-b ± √(b^2-4ac))/(2a).
Plugging in the coefficients (a=3, b=5, and c=-10) into the formula, we have y = (-(5) ± √((5)^2-4(3)(-10)))/(2(3)).
Calculate the discriminant (b^2-4ac) first, which is (5)^2 - 4(3)(-10) = 25 + 120 = 145.
Since the discriminant is positive, we know we will have two real solutions.
Proceed with the formula to find the two values of y: y = (-5 ± √145)/6.
Thus, we have two solutions for y, which are y = (-5 + √145)/6
= 1.173 and
y = (-5 - √145)/6.
= -2.84