Answer:
[tex]n = 3, n=-\frac92; t=9, t=-9[/tex]
Step-by-step explanation:
Second one is faster, so let's get rid of it first.
Divide by 2 to make numbers easier, and we get
[tex]t^2-81 = 0[/tex]. You should recognize 81 is 9 squared, and both [tex]-9[/tex] and [tex]+9[/tex], when squared, give 81, we have our solution.
Now the first. I can't spot any quick trick to solve it, so quadratic formula it is. Let's rememer it: if
[tex]ax^2 +bx+c=0[/tex] then
[tex]x={{-b \pm \sqrt{b^2-4ac}}\over 2a[/tex]
Let's bring the first equation in the standard form and calculate the quantity over the square root (usually called with the greek letter delta, [tex]\Delta[/tex]) to the side.
[tex]2n^2+3n-27 =0\\\Delta = (3)^2 -4(2)(-27) = 9+(8)(3)(9) = 9(1+24)= 9 \cdot25 =(3\cdot5)^2[/tex]
now we can apply the formula:
[tex]n={{-3\pm15}\over4}[/tex] Let's split the two cases now
[tex]n= \frac{-3+15}4 = \frac{12}4=3\\n= \frac{-3-15}4 = \frac{-18}4 =-\frac92[/tex]
