using the quadratic formula to solve 4x² - 3x + 9 = 2x + 1, what are the values of x?\n$\frac{1pmsqrt{159}i}{…

using the quadratic formula to solve 4x² - 3x + 9 = 2x + 1, what are the values of x?\n$\frac{1pmsqrt{159}i}{8}$\n$\frac{5pmsqrt{153}i}{8}$\n$\frac{5pmsqrt{103}i}{8}$\n$\frac{1pmsqrt{153}}{8}$

using the quadratic formula to solve 4x² - 3x + 9 = 2x + 1, what are the values of x?\n$\frac{1pmsqrt{159}i}{8}$\n$\frac{5pmsqrt{153}i}{8}$\n$\frac{5pmsqrt{103}i}{8}$\n$\frac{1pmsqrt{153}}{8}$

Answer

Explanation:

Step1: Rearrange the equation

First, rewrite $4x^{2}-3x + 9=2x + 1$ in standard quadratic - form $ax^{2}+bx + c = 0$. $4x^{2}-3x-2x + 9 - 1=0$, which simplifies to $4x^{2}-5x + 8 = 0$. Here, $a = 4$, $b=-5$, and $c = 8$.

Step2: Apply the quadratic formula

The quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substitute $a = 4$, $b=-5$, and $c = 8$ into the formula: $x=\frac{-(-5)\pm\sqrt{(-5)^{2}-4\times4\times8}}{2\times4}=\frac{5\pm\sqrt{25 - 128}}{8}=\frac{5\pm\sqrt{-103}}{8}=\frac{5\pm\sqrt{103}i}{8}$

Answer:

$\frac{5\pm\sqrt{103}i}{8}$