SOLUTION: log5(x^2+x+4)=2

Question 166806: log5(x^2+x+4)=2
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
log5(x^2+x+4)=2
x^2+x+4 = 5^2
x^2+x+4 = 25
x^2+x-21 = 0
Since we can't factor, we must use the quadratic equation.
Doing so will yield:
x = {4.110, -5.110}
.

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=85 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 4.10977222864644, -5.10977222864644. Here's your graph:

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