\(x= \frac{-b \pm \sqrt{ b^{2} -4ac} }{2a}\); where a = 2, b = 8 and c = -3
\(x= \frac{-8 \pm \sqrt{ 8^{2} -4 \times 2 \times -3} }{2 \times 2} \\ x= \frac{-8 \pm \sqrt{ 64 +24} }{4} \\ x= \frac{-8 \pm \sqrt{ 88} }{4} \\ x= \frac{-8 \pm 2\sqrt{ 22} }{4} \\ x= \frac{-4 + \sqrt{ 22} }{2}\ or\ \frac{-4 - \sqrt{ 22} }{2}\)
The sum of the roots is\(\frac{-4 + \sqrt{ 22} }{2}+ \frac{-4 - \sqrt{ 22} }{2} \\ = \frac{-8}{2} = -4\)
OR
simply sum of roots = -b/a = -8/2 = -4
