Answer
$x=\left\{ \dfrac{-5-\sqrt{41}}{4},\dfrac{-5+\sqrt{41}}{4} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
2k^2+5k-2=0
,$ use first the properties of equality to express the equation in the form $x^2\pm bx=c.$ Once in this form, complete the square by adding $\left( \dfrac{b}{2} \right)^2$ to both sides of the equal sign. Then express the left side as a square of a binomial while simplify the right side. Then take the square root of both sides (Square Root Property) and use the properties of equality to isolate the variable.
$\bf{\text{Solution Details:}}$
Using the properties of equality, in the form $x^2+bx=c,$ the given equation is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{2k^2+5k-2}{2}=\dfrac{0}{2}
\\\\
k^2+\dfrac{5}{2}k-1=0
\\\\
k^2+\dfrac{5}{2}k=1
.\end{array}
In the equation above, $b=
\dfrac{5}{2}
.$ The expression $\left( \dfrac{b}{2} \right)^2,$ evaluates to
\begin{array}{l}\require{cancel}
\left( \dfrac{\dfrac{5}{2}}{2} \right)^2
\\\\=
\left( \dfrac{5}{2}\div{2} \right)^2
\\\\=
\left( \dfrac{5}{2}\cdot\dfrac{1}{2} \right)^2
\\\\=
\left( \dfrac{5}{4} \right)^2
\\\\=
\dfrac{25}{16}
.\end{array}
Adding the value of $\left( \dfrac{b}{2} \right)^2,$ to both sides of the equation above results to
\begin{array}{l}\require{cancel}
k^2+\dfrac{5}{2}k+\dfrac{25}{16}=1+\dfrac{25}{16}
\\\\
k^2+\dfrac{5}{2}k+\dfrac{25}{16}=\dfrac{16}{16}+\dfrac{25}{16}
\\\\
k^2+\dfrac{5}{2}k+\dfrac{25}{16}=\dfrac{41}{16}
.\end{array}
With the left side now a perfect square trinomial, the equation above is equivalent to
\begin{array}{l}\require{cancel}
\left( k+\dfrac{5}{4} \right)^2=\dfrac{41}{16}
.\end{array}
Taking the square root of both sides (Square Root Property), simplifying the radical and then isolating the variable, the equation above is equivalent to
\begin{array}{l}\require{cancel}
k+\dfrac{5}{4}=\pm\sqrt{\dfrac{41}{16}}
\\\\
k+\dfrac{5}{4}=\pm\sqrt{\dfrac{1}{16}\cdot41}
\\\\
k+\dfrac{5}{4}=\pm\dfrac{1}{4}\sqrt{41}
\\\\
k+\dfrac{5}{4}=\pm\dfrac{\sqrt{41}}{4}
\\\\
k=-\dfrac{5}{4}\pm\dfrac{\sqrt{41}}{4}
\\\\
k=\dfrac{-5\pm\sqrt{41}}{4}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
k=\dfrac{-5-\sqrt{41}}{4}
\\\\\text{OR}\\\\
k=\dfrac{-5+\sqrt{41}}{4}
.\end{array}
Hence, $
x=\left\{ \dfrac{-5-\sqrt{41}}{4},\dfrac{-5+\sqrt{41}}{4} \right\}
.$
