using the quadratic formula to solve $4x^{2}-3x + 9 = 2x + 1$, what are the values of $x$?\n$\frac{1pmsqrt{15…

using the quadratic formula to solve $4x^{2}-3x + 9 = 2x + 1$, 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 $4x^{2}-3x + 9 = 2x + 1$, 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: Simplify the equation

$$ \begin{align*} 4x^{2}-3x + 9&=2x + 1\ 4x^{2}-3x-2x+9 - 1&=0\ 4x^{2}-5x + 8&=0 \end{align*} $$ For a quadratic equation (ax^{2}+bx + c=0) ((a = 4), (b=-5), (c = 8)), the quadratic formula is (x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}).

Step2: Calculate the discriminant (\Delta=b^{2}-4ac)

$$ \begin{align*} \Delta&=(-5)^{2}-4\times4\times8\ &=25-128\ &=-103 \end{align*} $$

Step3: Substitute into the quadratic formula

$$ \begin{align*} x&=\frac{-(-5)\pm\sqrt{- 103}}{2\times4}\ &=\frac{5\pm\sqrt{103}i}{8} \end{align*} $$

Answer:

(\frac{5\pm\sqrt{103}i}{8}) (corresponding to the third option (\frac{5\pm\sqrt{103}i}{8}))