Question

How do you solve this quadratic equation to find the complex solutions 2x^2+5=6x

Answer 1

`The\text{ solutions: }x=\frac{\text{3+{\imaginaryI}}}{2}\text{{\text{ or}} }\frac{\text{3-{\imaginaryI}}}{2}`

Step-by-step explanation:
`\begin{gathered} Given: \\ 2x^2+5=6x \end{gathered}`

We will rewrite the equation and solve for x:

`\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}`

Solve for x using quadratic function:

`\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{ = }(-(-6)\pm√((-6)^2-4(2)(5)))/(2*2) \end{gathered}`

`\begin{gathered} x\text{ = }(6\pm2\imaginaryI)/(4)=\text{ }(2(3\pm\imaginaryI))/(2(2))\text{ }(2\text{ was factorized out since it is common to denominator and numerator}) \\ x\text{ = }((3\pm\imaginaryI))/(2)\text{ }(2\text{ cancels in the numerator and denominator}) \\ x\text{ = }(3\pm\imaginaryI)/(2) \\ x\text{ = }\frac{3+\text{ }\mathrm{i}}{2}\text{ or }\frac{3-\text{ }\mathrm{i}}{2}\text{ } \end{gathered}`
`The\text{ solutions: }x=\frac{\text{3+{\imaginaryI}}}{2}\text{{\text{ or}} }\frac{\text{3-{\imaginaryI}}}{2}`