Step 1: If the sum and product of the zeroes are known, a quadratic polynomial is given by
$$p(x)=x^{2}-(\text{sum of zeroes})x+(\text{product of zeroes})$$
Step 2: Substitute sum $=-10$ and product $=24$.
$$p(x)=x^{2}-(-10)x+24=x^{2}+10x+24$$
Step 3: To find the zeroes, put $p(x)=0$ and factorise by splitting the middle term ($10=6+4$ and $6\times 4=24$).
$$x^{2}+6x+4x+24=0$$
Step 4: Group the terms.
$$x(x+6)+4(x+6)=0 \;\Rightarrow\; (x+6)(x+4)=0$$
Step 5: Therefore
$$x=-6 \quad\text{or}\quad x=-4$$
Step 6: Verification: sum $=-6+(-4)=-10$ ✓ and product $=(-6)(-4)=24$ ✓
Hence the polynomial is $x^{2}+10x+24$ and its zeroes are $-6$ and $-4$.