Respuesta :
[tex]The\text{ solutions: }x=\frac{\text{3+{\imaginaryI}}}{2}\text{{\text{ or}} }\frac{\text{3-{\imaginaryI}}}{2}[/tex]Explanation:[tex]\begin{gathered} Given: \\ 2x^2+5=6x \end{gathered}[/tex]
We will rewrite the equation and solve for x:
[tex]\begin{gathered} subtract\text{ 6x from both sides:} \\ 2x^2\text{ + 5 - 6x = 6x - 6x} \\ 2x^2\text{ + 5 - 6x = 0} \\ 2x^2\text{ - 6x + 5 = 0} \end{gathered}[/tex]Solve for x using quadratic function:
[tex]\begin{gathered} $x\text{ = }\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$ \\ 2x^2\text{ - 6x + 5 = 0} \\ \text{a = 2, b = -6, c = 5} \\ x\text{ = }\frac{-(-6)\pm\sqrt{(-6)^2-4(2)(5)}}{2\times2} \end{gathered}[/tex][tex]\begin{gathered} x=\frac{\text{-\lparen-6\rparen\pm}\sqrt{\text{\lparen-6\rparen}^2\text{ {\text{{}}}- 4\lparen2\rparen\lparen5\rparen}}}{\text{2\times2}} \\ x\text{ = }\frac{6\pm\sqrt{-4}}{4}\text{ } \\ Since\text{ we can't find square root of a negative number, we will use complex number} \\ In\text{ complex number, }i^2\text{ = -1} \\ substitute\text{ for -1 in the formula:} \\ \text{ x = }\frac{6\pm\sqrt{i^2\times4}}{4} \\ \text{ x=}\frac{6\pm i\sqrt{4}}{4}=\frac{6\pm2i}{4} \end{gathered}[/tex][tex]\begin{gathered} x\text{ = }\frac{6\pm2\imaginaryI}{4}=\text{ }\frac{2(3\pm\imaginaryI)}{2(2)}\text{ }(2\text{ was factorized out since it is common to denominator and numerator}) \\ x\text{ = }\frac{(3\pm\imaginaryI)}{2}\text{ }(2\text{ cancels in the numerator and denominator}) \\ x\text{ = }\frac{3\pm\imaginaryI}{2} \\ x\text{ = }\frac{3+\text{ }\mathrm{i}}{2}\text{ or }\frac{3-\text{ }\mathrm{i}}{2}\text{ } \end{gathered}[/tex][tex]The\text{ solutions: }x=\frac{\text{3+{\imaginaryI}}}{2}\text{{\text{ or}} }\frac{\text{3-{\imaginaryI}}}{2}[/tex]