\(\sqrt{25t^2-9}=2\sqrt{5t-3}\)

\(\sqrt{-2x^2+6}=x-1\)

help me

\(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}-\sqrt{2}\right)\)

\(\left(1+\dfrac{5-\sqrt{5}}{1-\sqrt{5}}\right)\left(\dfrac{5+\sqrt{5}}{1+\sqrt{5}}+1\right)\)

\(\left(\dfrac{1}{\sqrt{3}-\sqrt{2}}\right)\left(\dfrac{1}{\sqrt{3}-\sqrt{2}}\right)\)

\(\sqrt{25^2-24^2}\)

\(16x^2-8x+1\)=0

\(16x^2-4x-4x+1=0\)

\(\left(16x^2-4x\right)-\left(4x-1\right)=0\)

\(4x\left(4x-1\right)-\left(4x-1\right)=0\)

\(4x-1=0\)

4x=1

x= \(\dfrac{1}{4}\)