Answer :
Answer:
The roots are
[tex]x_1=-1+\sqrt{7}[/tex]
[tex]x_2=-1-\sqrt{7}[/tex]
Step-by-step explanation:
we know that
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^{2} +2x-6=0[/tex]
so
[tex]a=1\\b=2\\c=-6[/tex]
substitute in the formula
[tex]x=\frac{-2(+/-)\sqrt{2^{2}-4(1)(-6)}} {2(1)}[/tex]
[tex]x=\frac{-2(+/-)\sqrt{28}} {2}[/tex]
[tex]x=\frac{-2(+/-)2\sqrt{7}} {2}[/tex]
[tex]x_1=\frac{-2(+)2\sqrt{7}} {2}=-1+\sqrt{7}[/tex]
[tex]x_2=\frac{-2(-)2\sqrt{7}} {2}=-1-\sqrt{7}[/tex]
therefore
The roots are
[tex]x_1=-1+\sqrt{7}[/tex]
[tex]x_2=-1-\sqrt{7}[/tex]
Answer:
The roots are
what the dude above me said
Step-by-step explanation: A P E X