14. \( 3 x^{2}-35 x+60=x-21 \)

To solve the equation \( 3 x^{2}-35 x+60=x-21 \), we start by rearranging it to set it to zero. We subtract \( x \) and add \( 21 \) to both sides: \[ 3 x^{2} - 35 x + 60 - x + 21 = 0 \] This simplifies to: \[ 3 x^{2} - 36 x + 81 = 0 \] Next, we can divide the entire equation by 3 to simplify it further: \[ x^{2} - 12 x + 27 = 0 \] Now we use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a=1, b=-12, c=27 \): Calculating the discriminant: \[ b^2 - 4ac = (-12)^2 - 4(1)(27) = 144 - 108 = 36 \] Now substituting into the quadratic formula: \[ x = \frac{12 \pm \sqrt{36}}{2 \cdot 1} = \frac{12 \pm 6}{2} \] This gives us: \[ x = \frac{18}{2} = 9 \quad \text{and} \quad x = \frac{6}{2} = 3 \] Thus, the solutions are \( x = 9 \) and \( x = 3 \). Now you can easily check these solutions by substituting them back into the original equation!