Answer:[tex]x=-2+or-\sqrt{7}[/tex]
Step-by-step explanation:
[tex]x^{2} +4x-3=0[/tex]
[tex]x=\frac{-b+or-\sqrt{b^{2} -4ac} }{2a}[/tex]
[tex]x=\frac{-4 + or - \sqrt{4^{2}-4(1)(-3) } }{2(1)\\}\\x= -4+or-\sqrt{16+12}/2\\x=-4+or-\sqrt{28} /2\\x=-4+or-\sqrt{4*7} /2\\x=-4+or- 2\sqrt{7} /2\\x=-2+or-\sqrt{7}[/tex]
