Resposta:
[tex]\textsf{Leia abaixo}[/tex]
Explicação passo a passo:
[tex]\mathsf{x^2 - 3x - 10 = 10}[/tex]
[tex]\mathsf{x^2 - 3x - 20 = 0}[/tex]
[tex]\mathsf{\Delta = b^2 - 4.a.c}[/tex]
[tex]\mathsf{\Delta = (-3)^2 - 4.1.(-20)}[/tex]
[tex]\mathsf{\Delta = 9 + 80}[/tex]
[tex]\mathsf{\Delta = 89}[/tex]
[tex]\mathsf{x = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{3 \pm \sqrt{89}}{2} \rightarrow \begin{cases}\mathsf{x' = \dfrac{3 + \sqrt{89}}{2}}\\\\\mathsf{x'' = \dfrac{3 - \sqrt{89}}{2}}\end{cases}}[/tex]
[tex]\boxed{\boxed{\mathsf{S = \{\dfrac{3 + \sqrt{89}}{2};\dfrac{3 - \sqrt{89}}{2}\}}}}[/tex]
[tex]\mathsf{2x^2 + 4x + 3 = 0}[/tex]
[tex]\mathsf{\Delta = 4^2 - 4.2.3}[/tex]
[tex]\mathsf{\Delta = 16 - 24}[/tex]
[tex]\boxed{\boxed{\mathsf{\Delta = -8}}}\rightarrow\textsf{n{\~a}o possui ra{\'i}zes em }\mathbb{R}[/tex]
[tex]\mathsf{x = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{-4 \pm \sqrt{-8}}{2} \rightarrow \begin{cases}\mathsf{x' = \dfrac{-4 + 2\sqrt{2}\:i}{2} = -2 + \sqrt{2}\:i}\\\\\mathsf{x'' = \dfrac{-4 - 2\sqrt{2}\:i}{2} = -2 - \sqrt{2}\:i}\end{cases}}[/tex]
[tex]\boxed{\boxed{\mathsf{S = \{-2 + \sqrt{2}\:i;-2 - \sqrt{2}\:i\}}}}[/tex]
