The roots of \(7x^2+x-5=5x^2-3x\) are \(a\) and \(b\). Compute \((a-4)(b-4)\).
Move all the terms to the left side to get \(2x^2+4x-5=0\). Using the quadratic formula, we find that the roots are \(\frac{-4+\sqrt{16-4\cdot(-5)\cdot2}}{4}=\frac{-2+\sqrt{14}}{2}\) and \(\frac{-4-\sqrt{16-4\cdot(-5)\cdot2}}{4}=\frac{-2-\sqrt{14}}{2}\). So then, \((a-4)(b-4)\) is equal to \(43/2=21.5\).
