x^2=8x-3=0?

If your equation is x^2 - 8x - 3 = 0

Then

(x - 4)^2 - 16 - 3 = 0

(x - 4)^2 = 19

x = 4 ± √(19) where a = 4, b = 1 and C = 19


If your equations is x^2 + 8x - 3 = 0

Then

(x + 4)^2 - 16 - 3 = 0


(x + 4)^2 = 19

x = - 4 ± √(19), where a = -4, b = 1 and c = 19

YOU DON'T KNOW HOW TO PRESENT AN EQUATION !!

Read your problem AGAIN...then post it CORRECTLY !

x² - 8x + 3 = 0
x = [- b ± √ (b² - 4ac) ] / 2a
x = [ 8 ± √ (64 - 12) ] / 2
x = [ 8 ± √52 ] / 2
x = [ 8 ± 2√13 ] / 2
x = 4 ± √13

I think you mean x² = 8x-3, not x² = 8x-3 = 0.

x² = 8x-3
x² - 8x + 3 = 0

quadratic formula
x = [8 ± √(8² – 4·1·3)] / [2·1]
 = [8 ± √52] / 2
 = [8 ± 2√13] / 2
 = 4 ± √13

x² - 8x - 3 = 0

x² - 8x = 3

x² - 8x + 16 = 3 + 16

x² - 8x + 16 = 19

(x - 4)² = (± √19)²

x - 4 = ± √19

x = 4 ± √19 → compare it to: a + b√c

c = 19

b = 1

a = 4