Answer: A) x=-8±5[tex]\sqrt{2}[/tex]
Step-by-step explanation: to find the solutions of the given equation, we need to use the substitution u=x+2, so the equation would now be:
[tex]u^{2} +12u-14=0[/tex]
the quadratic formula is:
u=(-b±[tex]\sqrt{b^{2}-4ac }[/tex])/2a
in this case a=1; b=12; and c=-14
so replacing the values it remains:
u=(-12±[tex]\sqrt{12^{2}-4*1*(-14) }[/tex])/2*1
u=(-12±[tex]\sqrt{144+56}[/tex])/2
u=(-12±[tex]\sqrt{200}[/tex])/2
we can write 200 as 100*2 and the square root of 100 is 10:
u=(-12±10[tex]\sqrt{2}[/tex])/2
u=-6±5[tex]\sqrt{2}[/tex]
and finally:
x=u-2
x=-6±5[tex]\sqrt{2}[/tex]-2
x=-8±5[tex]\sqrt{2}[/tex]
