Verified Answer :

MattD

\(ax^2+bx+c=0\\\)
\(3x^2-7x+9=0\)
a=3 b=-7 c=9
\(\Delta=b^2-4ac=(-7)^2-4*3*9=49-108=-59\)
\(\Delta<0\)
If the delta is less than zero and if no polynomial real roots

Lilith

\(3x^2 - 7x + 9 = 0 \\ \\a=3, \ \ b=-7, \ \ c= 9 \\ \\ \Delta =b^2-4ac = 7^2 -4\cdot 3 \cdot 9 = 49-108= -59 \\ \\ If \ \Delta <0, \ then \ roots \ are \ imaginary \ (non-real)\)


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