#NOTE:\(e=equation\)

\(e\left(2\right)\Leftrightarrow y=3-3x\)

\(\Rightarrow e\left(1\right)\Leftrightarrow32x^3-48x^2+30x+\left(4\left(3-3x\right)-7\right)\sqrt{1-\left(3-3x\right)}=7\)

\(\Leftrightarrow32x^3-48x^2+30x+\left(5-12x\right)\sqrt{3x-2}=7\)

\(\Leftrightarrow32x^3-48x^2+10x+6+\left(5-12x\right)\sqrt{3x-2}-\left(-20x+13\right)=0\)

\(\Leftrightarrow2\left(x-1\right)\left(4x-3\right)\left(4x+1\right)+\dfrac{\left(x-1\right)\left(4x-3\right)\left(108x-73\right)}{\left(5-12x\right)\sqrt{3x-2}-20x+13}=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x-3\right)\left(2\left(4x+1\right)+\dfrac{\left(108x-73\right)}{\left(5-12x\right)\sqrt{3x-2}-20x+13}\right)=0\)

See: \(2\left(4x+1\right)+\dfrac{\left(108x-73\right)}{\left(5-12x\right)\sqrt{3x-2}-20x+13}>0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{4}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}y=3-3x=3-3\cdot1=0\\y=3-3x=3-3\cdot\dfrac{3}{4}=\dfrac{3}{4}\end{matrix}\right.\)

Lightning Farron:I don't understand

( ͡° ͜ʖ ͡°) 

Lightning Farron:\(2\left(4x+1\right)+\dfrac{108x-73}{\left(5-12x\right)\sqrt{3x-2}-20x+13}>0\).Why?