Solve the equation 3x^2 + 5x - 6 = 0.

Step-by-step solution

1. 1

   Identify the coefficients. Reading off the standard form $ax^{2}+bx+c=0$:

   $a=3,\qquad b=5,\qquad c=-6$

   Note $c=-6$ keeps its minus sign — carrying it correctly into the discriminant is the crux of the problem.

2. 2

   Compute the discriminant first.

   $\Delta=b^{2}-4ac=5^{2}-4(3)(-6)=25+72=97$

   Two negatives make the $-4ac$ term positive $72$. Since $\Delta=97>0$ there are two distinct real roots, but $97$ is prime — so it is not a perfect square, the roots are irrational, and no integer factorisation of $3x^{2}+5x-6$ exists.

3. 3

   Substitute into the quadratic formula.

   $x=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{-5\pm\sqrt{97}}{2(3)}=\frac{-5\pm\sqrt{97}}{6}$

4. 4

   Check that nothing simplifies further. $97$ has no square factors, so $\sqrt{97}$ cannot be reduced; and $-5$, $1$ and $6$ share no common factor, so the fraction stands as it is. The two roots are

   $x=\frac{-5+\sqrt{97}}{6}\approx0.808143,\qquad x=\frac{-5-\sqrt{97}}{6}\approx-2.474810$

5. 5

   Verify with Vieta and by substitution. The roots should sum to $-b/a=-\tfrac53$: $\dfrac{-5+\sqrt{97}}{6}+\dfrac{-5-\sqrt{97}}{6}=\dfrac{-10}{6}=-\dfrac53$ ✓. They should multiply to $c/a=-2$: $\dfrac{25-97}{36}=\dfrac{-72}{36}=-2$ ✓. Substituting either root into $3x^{2}+5x-6$ gives a residual below $10^{-11}$ ✓.