Solve equation
(t+1)^2 - 2(t-1)^2 = 6t-5
x^2-a(3x-2a+b)-b =0
I got no solution for both, is it right?
No.
$\displaystyle
t^2+2t+1 - 2(t^2-2t+1) - 6t + 5 = 0$
$\displaystyle -t^2 + 4 = 0$
(use the difference of two squares to solve)
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$\displaystyle x^2 - 3xa + 2a^2 - ab - b = 0$
Assuming $\displaystyle a \, , \, b \neq \, f(x)$
$\displaystyle x^2 - 3xa + (2a^2-ab-b) = 0$
(Use the quadratic formula to solve in terms of a and b)