Answer
two unequal real solutions
Work Step by Step
RECALL
The nature of solutions using the quadratic equation $ax^2+bx+c=0$ can be determined using the value of its discriminant $b^2-4ac$.
If the value of the discriminant is:
(1) negative, then the equation has no real solutions;
(2) zero, then the equation has one repeated real solution; and
(3) positive, then there are two unequal real solutions.
The given quadratic equation has :
$a=3
\\b=5
\\c=-8$
Solve for the discriminant to obtain:
$=b^2-4ac
\\=5^2-4(3)(-8)
\\=25-(-96)
\\=25+96
\\=121$
The discriminant is positive; therefore the equation has two unequal real solutions.
