Answer:
c = ± [tex]\sqrt{\frac{d+dt^2}{t^2-3} }[/tex]
Step-by-step explanation:
Given
t = [tex]\sqrt{\frac{3c^2+d}{c^2-d} }[/tex] ( square both sides )
t² = [tex]\frac{3c^2+d}{c^2-d}[/tex] ( multiply both sides by c² - d )
t²(c² - d) = 3c² + d ← distribute left side
t²c² - dt² = 3c² + d ( subtract 3c² from both sides )
t²c² - 3c² - dt² = d ( add dt² to both sides )
t²c² - 3c² = d +dt² ← factor out c² from each term on the left side
c²(t² - 3) = d + dt² ( divide both sides by t² - 3 )
c² = [tex]\frac{d+dt^2}{t^2-3}[/tex] ( take the square root of both sides )
c = ± [tex]\sqrt{\frac{d+dt^2}{t^2-3} }[/tex]
