Math help...5x^2+x-1=0

5x^2+x-1=0

-1+√(21)/10 or -1-√(21)/10

3 ways to solve this...
BUT the EASIEST is by using the quadratic formula. Look it up on google for a better picture of it:
This is the quadratic formula:

x= -b +/- [(b^2-4ac)/2a] ^ 1/2 Note that the ^1/2 means the whole thing in the brackets is under a square root. I don't have the sqrt sign so I used the other form of expressing it which is ^1/2
It's like: The square root of x, is the same as x^1/2

therefore we are looking for 3 things: a, b, and c
a= the number before x^2
b= the number before x
c= the single number, AKA the constant (number without letter)

In this case a=5, b=1, c=-1
therefore using the formula and plugging in the values:

x= +/-.....
Oh one more thing...+/- means x is plus OR minus meaning you can seperate into two equations for two values of x.
BUT just to save space and time: we put it as +/-

So...
x=+/- [(1)^2 - 4(5)(-1)]^1/2 /2(5)
x=+/- [1+20]^1/2 /10
x=+/- square root of 21, which is approximately 4.58/10
so we split the equation into two seperate ones.
x=+4.58/10 or x=-4.58/10


x=0.458 or x= -0.458


Hope it helped!

using the quadratic formula:

x = [-1 +/- sqrt(1^2 - 4(5)(-1))]/2(5)
= [-1 +/- sqrt(1+20)] / 10
= [-1 +/- 4.58] / 10
= 0.358 or -0.558

according to the quadratic formula x=-0.689 and 0.289, is that what you're looking for?

5x^2 + x - 1 = 0
a = 5; b = 1; c = -1
x
= [-1 +/- sqrt (1 + 20)]/10
= (-1 +/- sqrt 21)/10

=4?