Solve for :
Try integer factoring first, and watch it fail. Factoring needs two integers whose product is and whose sum is . Run through the factor pairs of : sums to , sums to , sums to , sums to , sums to . Nothing lands on , so no integer factorization exists. A frequently repeated wrong answer here is — but that expands to , not .
Confirm the failure with the discriminant. For with , , ,
Since and , the value sits strictly between two consecutive squares, so it is not a perfect square. That is exactly the algebraic reason the integer search above could never succeed: rational roots require a square discriminant.
Apply the quadratic formula. With there are two distinct irrational roots:
This is the exact answer; is prime-free of square factors, so cannot be simplified further.
Convert to decimals. , so
Note how far is from the tempting : the two are not interchangeable.
Check with Vieta's formulas. The sum of the roots must be and the product must be . Indeed , and exactly,
Both conditions hold, so the roots are correct.
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