Answer:
x = -5 , 3
Step-by-step explanation:
(x₁ , y₁) = (-1, -2) & (x₂, y₂) = (x , 2)
[tex]Distance =\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\\\\\\\sqrt{(x - [-1])^{2}+(2-[-2])^{2}}=2\sqrt{5}\\\\\sqrt{(x+1)^{2}+(2+2)^{2}}=2\sqrt{5}\\\\\sqrt{(x)^{2}+2x+1+(4)^{2}}=2\sqrt{5}\\\\\sqrt{x^{2}+2x+1+4}=2\sqrt{5}\\\\\sqrt{x^{2}+2x+5}=2\sqrt{5}[/tex]
Take square both sides,
x² + 2x + 5 = (2)²(√5)² {(√5)² = √5*√5 = 5 }
x² + 2x +5 = 4 * 5
x² + 2x + 5 = 20
x² + 2x + 5 - 20 = 0
x² + 2x -15 = 0
x² + 5x - 3x - 5*3 = 0
x(x + 5) -3(x + 5)=0
(x + 5)(x - 3) = 0
x + 5 = 0 ; x - 3 = 0
x = -5 ; x = 3
x = -5 , 3
