Question

solve for x in the equation $2x^2 - 5x + 1 = 3$. $\bigcirc$ $x = \frac{5}{2} \pm \frac{\sqrt{29}}{2}$ $\bigcirc$ $x = \frac{5}{2} \pm \frac{\sqrt{41}}{4}$ $\bigcirc$ $x = \frac{5}{4} \pm \frac{\sqrt{29}}{2}$ $\bigcirc$ $x = \frac{5}{4} \pm \frac{\sqrt{41}}{4}$

Explanation:

Step1: Rewrite to standard quadratic form

Subtract 3 from both sides:
$2x^2 -5x +1 -3 = 0$
$2x^2 -5x -2 = 0$

Step2: Identify coefficients

For $ax^2+bx+c=0$, $a=2$, $b=-5$, $c=-2$

Step3: Apply quadratic formula

Quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Substitute values:
$x=\frac{-(-5)\pm\sqrt{(-5)^2-4(2)(-2)}}{2(2)}$

Step4: Simplify the expression

Calculate discriminant:
$\sqrt{25 + 16}=\sqrt{41}$
Simplify numerator and denominator:
$x=\frac{5\pm\sqrt{41}}{4}$

Answer:

$x=\frac{5}{4}\pm\frac{\sqrt{41}}{4}$