Question

How can I solve the following quadratic equations with the quadratic formula?

a) x^2 + 5x + 6 = 0

b) 2x^2 - 3x - 2 = 0

Answer 1

`~~~~~~~~~~~~\textit{quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{1}x^2\stackrel{\stackrel{b}{\downarrow }}{+5}x\stackrel{\stackrel{c}{\downarrow }}{+6}=0 \qquad \qquad x= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ x= \cfrac{ - (5) \pm \sqrt { (5)^2 -4(1)(6)}}{2(1)} \implies x = \cfrac{ -5 \pm \sqrt { 25 -24}}{ 2 } \\\\\\ x= \cfrac{ -5 \pm \sqrt { 1 }}{ 2 }\implies x=\cfrac{-5\pm 1}{2}\implies x= \begin{cases} -2\\ -3 \end{cases} \\\\[-0.35em] ~\dotfill`

`~~~~~~~~~~~~\textit{quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{2}x^2\stackrel{\stackrel{b}{\downarrow }}{-3}x\stackrel{\stackrel{c}{\downarrow }}{-2}=0 \qquad \qquad x= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ x= \cfrac{ - (-3) \pm \sqrt { (-3)^2 -4(2)(-2)}}{2(2)} \implies x = \cfrac{ 3 \pm \sqrt { 9 +16}}{ 4 } \\\\\\ x= \cfrac{ 3 \pm \sqrt { 25 }}{ 4 }\implies x=\cfrac{3\pm 5}{4}\implies x= \begin{cases} 2\\ -(1)/(2) \end{cases}`