what are the solutions and line of symmetry for x^2+4x-5=0?

Question

Using the graph of y=x^2+4x-5,
1)what are the solutions to the equations x^2+4x-5=0?
and 2)equation of line of symmetry for this?

Answer 1

Part 1
( x + 5 ) ( x - 1 ) = 0
x = - 5 , x = 1

Part 2
( x ² + 4 x + 4 ) - 4 - 5 = 0
( x + 2 ) ² = 9
x = - 2 is line of symmetry

Answer 2

We know the -5 is the product of two numbers. These same two numbers when each is multiplied by x will sum to 4x. 5-1=4 and -1 times 5 is -5. So the two numbers are 5 and -1
(x+5)(x-1)=x^2 -x +5x -5 = x^2+4x-5. We know this is a parabola and that it is zero at x=-5 and x=1. That is 6 from x intercept to x intercept so the center is 3 away from the intercepts which is the vertical line x=-2. We could also find the top of the parabola by finding where the first derivative is zero => 2x+4=0 at x=-2 so it checks. Line of symmetry is x=-2