i need algebra help final in 2 hrs 1) y=x^2+4x-5?

Question

i need algebra help final in 2 hrs
1.) y=x^2+4x-5
2.) x^2-12x+36=-8
3.)(2x+1)^2+(2x-1)^2=10
please help me i cannot afford to fail this class thank you very much in advance

it says solve the equation
1.) y=x^2+4x-5
2.) x^2-12x+36=-8

Give the X-intercept and the coordinates of the vertex

3.)(2x+1)^2+(2x-1)^2=10

Answer 1

1.) y=x^2+4x-5
y = (x+5)(x-1)
if y = 0, then x = -5 or x = 1 [They make the value in the parentheses = 0 and henc make y =0]

2.) x^2-12x+36=-8
(x-6)^2 = -8
x-6 = +/- sqrt(-8) {take sqrt of both sides}
x = 6 +/- sqrt -8)
x = 6 +/- 2isqrt(2) {i is the sqrt of -1. The answers are imaginary}

3.)(2x+1)^2+(2x-1)^2=10
4x^2 +4x + 1 +4x^2 -4x +1 = 10
8x^2 +2 = 10
8x^2 = 8
x^2 =1
x = +/- 1

Answer 2

I'm not sure what you're trying to do here for #1
If you're factoring, then
x^2 + 4x - 5 = (x + 5)(x - 1)

#2
x^2 - 12x + 36 = - 8
(x - 6)(x - 6) = -8

#3
(2x + 1)^2 + (2x - 1)^2 = 10
Not sure what you want to do here.

Answer 3

All you did was give equations, but you never said what you are trying to solve for or anything.

Answer 4

Use the quadratic formula,