Solve the quadratic equation 40x2 + 98x - 33 = 0.

Step-by-step solution

1. 1

   Compute the discriminant before anything else. With $a=40$, $b=98$, $c=-33$,

   $\Delta=b^2-4ac=98^2-4(40)(-33)=9604+5280=14884$

   A leading coefficient of $40$ makes trial-and-error factoring painful, so it is worth spending one line on $\Delta$ to find out whether nice factors even exist.

2. 2

   Recognise 14884 as a perfect square. Strip the obvious factor of $4$:

   $14884=4\cdot 3721,\qquad \sqrt{14884}=2\sqrt{3721}=2\cdot 61=122$

   because $61^2=3721$. A square discriminant guarantees two rational roots, which in turn guarantees a clean factorization over the integers.

3. 3

   Substitute into the quadratic formula.

   $x=\frac{-98\pm 122}{2\cdot 40}=\frac{-98\pm 122}{80}$

   Both branches now reduce to simple fractions rather than decimals.

4. 4

   Reduce each root.

   $x=\frac{-98+122}{80}=\frac{24}{80}=\frac{3}{10},\qquad x=\frac{-98-122}{80}=\frac{-220}{80}=-\frac{11}{4}$

   Always reduce: $24/80$ and $-220/80$ both hide a common factor.

5. 5

   Read off the factorization and verify. Roots $3/10$ and $-11/4$ correspond to factors $(10x-3)$ and $(4x+11)$, and their leading coefficients multiply to $10\cdot 4=40$:

   $(10x-3)(4x+11)=40x^2+110x-12x-33=40x^2+98x-33$

   The middle term rebuilds to $98x$, so both the roots and the factored form are confirmed.