(1)
we are given
[tex]x^2+6x+10=0[/tex]
we can use quadratic formula
[tex]\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}[/tex]
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
now, we can compare and find a , b and c
we get
[tex]a=1,b=6,c=10[/tex]
now, we can plug these values into quadratic formula
[tex]x=\frac{-6\pm \sqrt{6^2-4(1)(10)}}{2(1)}[/tex]
[tex]x=\frac{-6+\sqrt{6^2-4\cdot \:1\cdot \:10}}{2\cdot \:1}[/tex]
we can simplify it
[tex]x=-3+i[/tex]
[tex]x=\frac{-6-\sqrt{6^2-4\cdot \:1\cdot \:10}}{2\cdot \:1}[/tex]
[tex]x=-3-i[/tex]
so, we will get solution
{−3+i, −3−i}.........Answer
(2)
we are given equation as
[tex]x^2-8x+14=0[/tex]
Since, Jamal solve this equation by completing square
so, firstly we will move constant term on right side
so, subtract both sides by 14
[tex]x^2-8x+14-14=0-14[/tex]
[tex]x^2-8x=-14[/tex]
we can write
[tex]-8x=-2\times 4\times x[/tex]
so, we will add both sides by 4^2
[tex]x^2-8x+4^2=-14+4^2[/tex]
we get
[tex]x^2-8x+16=-14+16[/tex]..............Answer