## Algebra 2 Common Core

$\left\{-\dfrac{5}{4}-\dfrac{\sqrt{37}}{4}, -\dfrac{5}{4}+\dfrac{\sqrt{37}}{4}\right\}$
Add $3$ to both sides to obtain: $$4x^2+10x=3$$ Divide $4$ to both sides: \begin{align*} x^2+\frac{10x}{4}&=\frac{3}{4}\\\\ x^2+\frac{5}{2}x&=\frac{3}{4} \end{align*} RECALL: To complete the square for $x^2 +mb$, just add $\left(\dfrac{m}{2}\right)^2$. Use the rule above to complete the square (note that to retain the equality, whatever is added on the left side of the equation must also be added to the right side of the equation) and obtain: \begin{align*} x^2+\frac{5}{2}x+\left(\dfrac{\frac{5}{2}}{2}\right)^2&=\frac{3}{4}+\left(\dfrac{\frac{5}{2}}{2}\right)^2\\\\ x^2+\frac{5}{2}x+\left(\frac{5}{4}\right)^2&=\frac{3}{4}+\left(\frac{5}{4}\right)^2\\\\ x^2+\frac{5}{2}x+\frac{25}{16}&=\frac{3}{4}+\frac{25}{16}\\\\ \left(x+\frac{5}{4}\right)^2&=\frac{12}{16}+\frac{25}{16}\\\\ \left(x+\frac{5}{4}\right)^2&=\frac{37}{16}\\\\ \end{align*} Take the square root of both sides to obtain: \begin{align*} \sqrt{\left(x+\frac{5}{4}\right)^2}&=\pm\sqrt{\frac{37}{16}}\\\\ x+\frac{5}{4}&=\pm \dfrac{\sqrt{37}}{4}\\\\ x&=-\frac{5}{4}\pm \dfrac{\sqrt{37}}{4}\\\\ \end{align*} Thus, the solutions are: $\left\{-\dfrac{5}{4}-\dfrac{\sqrt{37}}{4}, -\dfrac{5}{4}+\dfrac{\sqrt{37}}{4}\right\}$