Question

1. $6x^2 - 16x - 6 = 2x^2 + 7x$

Explanation:

Step1: Move all terms to left side

Subtract \(2x^{2}\) and \(7x\) from both sides: \(6x^{2}-16x - 6-2x^{2}-7x=0\)
Simplify: \(4x^{2}-23x - 6 = 0\)

Step2: Factor the quadratic equation

We need two numbers \(a\) and \(b\) such that \(a\times b=4\times(- 6)=-24\) and \(a + b=-23\). The numbers are \(-24\) and \(1\).
Rewrite the middle term: \(4x^{2}-24x+x - 6 = 0\)
Factor by grouping: \(4x(x - 6)+1(x - 6)=0\)
\((4x + 1)(x - 6)=0\)

Step3: Solve for x

Set each factor equal to zero:
\(4x+1 = 0\) or \(x - 6 = 0\)
For \(4x+1 = 0\), \(4x=-1\), \(x=-\frac{1}{4}\)
For \(x - 6 = 0\), \(x = 6\)

Answer:

\(x = 6\) or \(x=-\frac{1}{4}\)