Solve the equation -t^2 - 6t + 28 = 0.

Step-by-step solution

1. 1

   Multiply the whole equation by $-1$. For an equation (unlike an inequality) this is free — it changes no roots, and a positive leading coefficient makes everything easier to read:

   $t^2+6t-28=0$

2. 2

   Try to factor, and record why it fails. We need integers with product $-28$ and sum $6$. The candidate pairs are $(1,-28),(2,-14),(4,-7),(7,-4),(14,-2),(28,-1)$, whose sums are $-27,-12,-3,3,12,27$. None is $6$, so no integer factorisation exists.

3. 3

   Compute the discriminant and apply the formula. With $a=1$, $b=6$, $c=-28$:

   $\Delta=36+112=148\quad\Longrightarrow\quad t=\frac{-6\pm\sqrt{148}}{2}$

4. 4

   Simplify. $148=4\cdot 37$ and $37$ is prime, so $\sqrt{148}=2\sqrt{37}$:

   $t=\frac{-6\pm 2\sqrt{37}}{2}=-3\pm\sqrt{37}$

5. 5

   Check numerically in the original equation. $\sqrt{37}\approx 6.0828$, so $t\approx 3.0828$ or $t\approx-9.0828$. Then $-(3.0828)^2-6(3.0828)+28\approx-9.503-18.497+28\approx 0$ $\checkmark$.