Answer:
[tex]d=\frac{5}{2},\\\\d=\frac{2}{5}[/tex]
Step-by-step explanation:
You can use the quadratic formula for any quadratic. The quadratic formula is given by [tex]\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex], and yields both real and non-real solutions.
In the given equation [tex]10d^2-29d+10=0[/tex], we have:
[tex]a=10,\\b=-29,\\c=10[/tex]
Substituting given values into the quadratic formula, we have:
[tex]x=\frac{29\pm\sqrt{(-29)^2-4(10)(10)}}{2(10)},\\\\x=\frac{29\pm\sqrt{441}}{20},\\\\x=\frac{29\pm 21}{20},\\\\\begin{cases}x=\frac{29+21}{20},x=\frac{50}{20}=\boxed{\frac{5}{2}}\\x=\frac{29-21}{20}, x=\frac{8}{20}=\boxed{\frac{2}{5}}\end{cases}[/tex]
The attempt in the problem is correct. Sometimes with auto-grading, the solutions need to be in a specific order to be marked correct. I'd suggest submitting the solutions in a different order to check if that's the issue.
