Respuesta :

Answer:

[tex]x=1+\frac{\sqrt[]{40}}{4}[/tex]

Explanation: Given the folllowing equation, we need its positive solution or root:

[tex]2x^2-4x-30=0[/tex]

Solution by Quadratic formula:

[tex]x=\frac{-B\pm\sqrt[]{B^6-4AC}}{2S}\rightarrow(1)[/tex]

Where x can be positive and negative, but we are interested in the positive solution.

Constants A B C in (1) are as follows:

[tex]\begin{gathered} A=2 \\ B=-4 \\ C=-30 \end{gathered}[/tex]

Plugging these values in (1) gives us:

[tex]x=\frac{-(-4)\pm\sqrt[]{(-4)^2-4(2)(-30)}}{2(2)}=\frac{4\pm\sqrt[]{16+(8)(30}}{4}[/tex]

Firther simplification gives:

[tex]\begin{gathered} x=\frac{4\pm\sqrt[]{16+24}}{4}=\frac{4\pm\sqrt[]{40}}{4}\rightarrow positive\rightarrow x=\frac{4+\sqrt[]{40}}{4}=1+\frac{\sqrt[]{40}}{4} \\ \therefore\rightarrow \\ x=1+\frac{\sqrt[]{40}}{4} \end{gathered}[/tex]

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