Question

2n²+3n=27

2t²-162=0
find the roots of each equation.​

Answer 1

`n = 3, n=-\frac92; t=9, t=-9`

Explanation:

Second one is faster, so let's get rid of it first.

Divide by 2 to make numbers easier, and we get

`t^2-81 = 0`. You should recognize 81 is 9 squared, and both
`-9` and
`+9`, 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

`ax^2 +bx+c=0` then

`x={{-b \pm √(b^2-4ac)}\over 2a`

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,
`\Delta`) to the side.

`2n^2+3n-27 =0\\\Delta = (3)^2 -4(2)(-27) = 9+(8)(3)(9) = 9(1+24)= 9 \cdot25 =(3\cdot5)^2`

now we can apply the formula:

`n={{-3\pm15}\over4}` Let's split the two cases now

`n= \frac{-3+15}4 = \frac{12}4=3\\n= \frac{-3-15}4 = \frac{-18}4 =-\frac92`