First we'll do two basic steps. Step 1 is to subtract 18 from both sides. After that, divide both sides by 2 to get x^2 all by itself. Let's do those two steps now
2x^2+18 = 10
2x^2+18-18 = 10-18 <<--- step 1
2x^2 = -8
(2x^2)/2 = -8/2 <<--- step 2
x^2 = -4
At this point, it should be fairly clear there are no solutions. How can we tell? By remembering that x^2 is never negative as long as x is real.
Using the rule that negative times negative is a positive value, it is impossible to square a real numbered value and get a negative result.
For example
2^2 = 2*2 = 4
8^2 = 8*8 = 64
(-10)^2 = (-10)*(-10) = 100
(-14)^2 = (-14)*(-14) = 196
No matter what value we pick, the result is positive. The only exception is that 0^2 = 0 is neither positive nor negative.
So x^2 = -4 has no real solutions. Taking the square root of both sides leads to
x^2 = -4
sqrt(x^2) = sqrt(-4)
|x| = sqrt(4)*sqrt(-1)
|x| = 2*i
x = 2i or x = -2i
which are complex non-real values
Oh ok. So I think that x is 83. Because I did this 630-100-121-106-112-108=83. And then I did 630/6 and got 105 and thats the mean.
If triangle ABC is equilateral,solve for X56x-308
8.8 because .78 rounded up is .8
