Question

20(x - 2)^2 - 5(x + 1)^2 + 48(x - 2)(x + 1) = 0

Answer 1

\(20\left(x-2\right)^2-5\left(x+1\right)^2+48\left(x-2\right)\left(x+1\right)=0\)

<=> \(63x^2-138x-21=0\)

<=> \(3\left(21x^2-46x-7\right)=0\)

<=> \(3\left(21x^2+3x-49x-7\right)=0\)

<=> \(3\left[3x\left(7x+1\right)-7\left(7x+1\right)\right]=0\)

<=> \(3\left(7x+1\right)\left(3x-7\right)=0\)

<=> \(\orbr{\begin{cases}7x+1=0\\3x-7=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=-\frac{1}{7}\\x=\frac{3}{7}\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-\frac{1}{7}\\x=\frac{3}{7}\end{cases}}\)

Answer 2

\(20\left(x-2\right)^2-5\left(x+1\right)^2+48\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow63x^2-138x-21=0\)

\(\Leftrightarrow3\left(7x+1\right)\left(3x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{7}\\x=\frac{7}{3}\end{cases}}\)