Question

Determine os valores reais de t para que se tenha ( 2t + 1 )² - 2(2t + 1 )=8

Answer 1

$(2t+1)^2-2(2t+1)=8\\ \\ 4t^2+4t+1-4t-2-8=0\\ \\ 4t^2-9=0\\ \\ 4t^2=9\\ \\ t^2=\frac{9}{4}\\ \\ t=\pm\sqrt{\frac{9}{4}}\\ \\ \boxed{t=\pm \frac{3}{2}}$

Answer 2

$(2t+1)^2-2(2t+1)=8$

$4t^2+4t+1-4t-2=8$

$4t^2=8+1=9$

$t^2=\dfrac{9}{4}$

$t=\pm\dfrac{3}{2}$