Answer:
The solution set is x={2,-12}
Step-by-step explanation:
Given: \(x^2+10x=24\)
Using complete square to solve for x
\(x^2+10x=24\)
Add both side square of half of coefficient of x ( i,e 25)
\(x^2+10x+25=24+25\)
\((x+5)^2=49\) \(\because a^2+2ab+b^2=(a+b)^2\)
Taking square root both sides
\(\sqrt{(x+5)^2}=\pm \sqrt{49}\)
\(x+5=\pm7\)
\(x=-5\pm7\)
\(x=-5+7\) and \(x=-5-7\)
\(x=2,-12\)
Hence, The solution set is x={2,-12}
