\(TH1\\ \left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\dfrac{3}{2}\right)^2\\ \dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\\ \dfrac{3}{2}x=\dfrac{1}{5}-\dfrac{3}{2}\\ \dfrac{3}{2}x=-\dfrac{13}{10}\\ x=-\dfrac{13}{15}\\ TH2\\ \left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(-\dfrac{3}{2}\right)^2\\ \dfrac{1}{5}-\dfrac{3}{2}x=-\dfrac{3}{2}\\ \dfrac{3}{2}x=\dfrac{1}{5}-\left(-\dfrac{3}{2}\right)\\ \dfrac{3}{2}x=\dfrac{17}{10}\\ x=\dfrac{17}{15}\)