When b²−4ac=0 there is one real root.
When b²−4ac>0 there are two real roots.
When b²−4ac<0 no real roots or two complex roots
First equation
-x²-4x+7
[tex]\begin{gathered} b^{2}-4ac \\ \mleft(-4\mright)^2-4\mleft(-1\mright)\cdot\: 7 \\ 16+28=44 \end{gathered}[/tex]b²−4ac>0, then equation -x²-4x+7 has two real roots.
Second equation
-2x²+9x-11
[tex]\begin{gathered} b^{2}-4ac \\ 9^2-4\mleft(-2\mright)\mleft(-11\mright) \\ 81-88=-7 \end{gathered}[/tex]b²−4ac<0, then equation -2x²+9x-11 has two complex roots.
[tex]x1=\frac{9}{4}-i\frac{\sqrt{7}}{4},\: x2=\frac{9}{4}+i\frac{\sqrt{7}}{4}[/tex]Third equation
3x²-6x+3
[tex]\begin{gathered} b^{2}-4ac \\ \mleft(-6\mright)^2-4\cdot\: \: 3\cdot\: \: 3 \\ 36-36=0 \end{gathered}[/tex]b²−4ac=0, then equation 3x²-6x+3 has one root.
