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 : $a=2 \\b=-3 \\c=-7$ Solve for the discriminant to obtain: $=b^2-4ac \\=(-3)^2-4(2)(-7) \\=9-(-56) \\=9+56 \\=65$ The discriminant is positive; therefore the equation has two unequal real solutions.