\(\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)
⇔ \(\dfrac{3\left(2x+1\right)^2}{15}-\dfrac{5\left(x-1\right)^2}{15}=\dfrac{7x^2-14x-5}{15}\)
⇔ \(3\left(2x+1\right)^2-5\left(x-1\right)^2=7x^2-14x-5\)
⇔ \(3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)=7x^2-14x-5\)
⇔ \(12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)
⇔ \(7x^2+22x-2=7x^2-14x-5\) ⇔ \(36x+3=0\) ⇔ x=\(\dfrac{-1}{12}\)
\(\Leftrightarrow3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)=7x^2-14x-5\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)
\(\Leftrightarrow36x=-3\)
hay x=-1/12
