Solve -t² - 6t + 28 = 0 for t, given that t is positive.

Step-by-step solution

1. 1

   Clear the negative leading coefficient. Multiplying an equation by $-1$ changes nothing about its solution set (unlike an inequality), and it makes every later sign easier:

   $-t^2-6t+28=0\quad\Longleftrightarrow\quad t^2+6t-28=0$

2. 2

   Compute the discriminant. With $a=1$, $b=6$, $c=-28$,

   $\Delta=6^2-4(1)(-28)=36+112=148$

   Since $\Delta>0$ there are two distinct real roots, and $148=4\cdot 37$ is not a perfect square, so they will be irrational.

3. 3

   Simplify the radical before dividing.

   $\sqrt{148}=\sqrt{4\cdot 37}=2\sqrt{37}$

   Pulling the $2$ out now is what lets the whole formula reduce cleanly in the next step.

4. 4

   Apply the quadratic formula and reduce.

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

   Both the $-6$ and the $2\sqrt{37}$ are divided by $2$ — halving only one of them is the usual error.

5. 5

   Use the condition $t>0$ to select the root. With $\sqrt{37}\approx 6.0828$:

   $-3+\sqrt{37}\approx 3.0828>0,\qquad -3-\sqrt{37}\approx -9.0828<0$

   Only the first survives. (Vieta agrees: the product of the roots is $-28<0$, so exactly one root is positive.)

6. 6

   Verify. With $t=-3+\sqrt{37}$, $t^2=9-6\sqrt{37}+37=46-6\sqrt{37}$, so

   $-t^2-6t+28=-46+6\sqrt{37}+18-6\sqrt{37}+28=0$

   exactly, confirming $t=-3+\sqrt{37}\approx 3.0828$.