Answer:
The roots of the given equation are -5 + 2i and -5 - 2i
Step-by-step explanation:
Since, roots of an equation are the points which satisfy the equation,
Or the points which are obtained after solving the equation. ( by putting zero on right side )
Here, the given quadratic equation,
\(x^2+10x+29=0\)
If we have an equation, ax² +bx + c = 0
By quadratic formula,
\(x=\frac{-b\pm \sqr{b^2-4ac}}{2a}\)
By comparing,
The solution of the given equation is,
\(x=\frac{-10\pm \sqrt{10^2-4\times 1\times 29}}{2\times 1}\)
\(x=\frac{-10\pm \sqrt{100-116}}{2}\)
\(x=\frac{-10\pm \sqrt{-16}}{2}\)
\(x=\frac{-10\pm +4i}{2}\)
\(\implies x = -5\pm 2i\)
Thus, the roots of the given equation are -5 + 2i and -5 - 2i
