Answer :
Hi, the quadratic formula is: [tex] \boxed{X= \dfrac{-b \pm \sqrt{b ^{2}-4ac} }{2a}}[/tex]
Values:
a = 1; b = - 2; c = - 3
SOLVING:
[tex]0=X ^{2}-2X-3[/tex]
[tex]X ^{2}-2X-3=0[/tex]
[tex]X= \dfrac{-(-2) \pm \sqrt{(-2) ^{2}-4(1)(-3) } }{2(1)}[/tex]
[tex]X= \dfrac{2 \pm \sqrt{4+12} }{2}[/tex]
[tex]X= \dfrac{2 \pm \sqrt{16} }{2}[/tex]
[tex]X= \dfrac{2 \pm4}{2}[/tex]
Getting X₁:
X₁ = [tex] \dfrac{2+4}{2}[/tex]
X₁ = [tex] \dfrac{6}{2}[/tex]
X₁ = [tex] \boxed{ \boxed{ \textbf{3}}}\ ===>\ The\ first\ solution.[/tex]
Getting X₂:
X₂ = [tex] \dfrac{2-4}{2}[/tex]
X₂ = [tex] \dfrac{-2}{2}[/tex]
X₂ = [tex] \boxed{ \boxed{ \textbf{-1}}}\ ===>\ The\ second\ solution.[/tex]
GOOD LUCK...!!!
Values:
a = 1; b = - 2; c = - 3
SOLVING:
[tex]0=X ^{2}-2X-3[/tex]
[tex]X ^{2}-2X-3=0[/tex]
[tex]X= \dfrac{-(-2) \pm \sqrt{(-2) ^{2}-4(1)(-3) } }{2(1)}[/tex]
[tex]X= \dfrac{2 \pm \sqrt{4+12} }{2}[/tex]
[tex]X= \dfrac{2 \pm \sqrt{16} }{2}[/tex]
[tex]X= \dfrac{2 \pm4}{2}[/tex]
Getting X₁:
X₁ = [tex] \dfrac{2+4}{2}[/tex]
X₁ = [tex] \dfrac{6}{2}[/tex]
X₁ = [tex] \boxed{ \boxed{ \textbf{3}}}\ ===>\ The\ first\ solution.[/tex]
Getting X₂:
X₂ = [tex] \dfrac{2-4}{2}[/tex]
X₂ = [tex] \dfrac{-2}{2}[/tex]
X₂ = [tex] \boxed{ \boxed{ \textbf{-1}}}\ ===>\ The\ second\ solution.[/tex]
GOOD LUCK...!!!
Answer:
its A
Step-by-step explanation:
guessed and got it right