What is the solution for 5x^2-x+6=0

The easiest way to solve this problem is by using the Pythagorean theorem :D

Pythagorean theorem: [-b +- sqrt(b^2 - 4ac)]/2a

5 = a

-1 = b

6 = c

therefore, by plugging these values in!

[-(-1) +- sqrt((-1)^2 - 4(5)(6))]/2(5)

[1 +- sqrt(1 - 120)]/10

Oh so we're using imaginary numbers, so in this case you'll need to know that i = sqrt(-1), so keep this in mind (I can see why this was so difficult)

[1 +- sqrt(119)*i]10

answers:

( 1 + i*sqrt(119) ) / 10

( 1 - i*sqrt(119) ) / 10

i can answer this question sweetie!

step-by-step explanation:

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!