Solve for \( x \) : \( \begin{array}{ll}\text { (1) } 4 x^{2}=9 x & \text { (2) } x^{2}=25 \\ \text { (3) } x^{2}-5 x=14 & \text { (4) }-x^{2}-4 x+12=0 \\ \text { (5) } 2 x^{2}-5 x+3=0 & \text { (6) } 6 x^{2}+7 x=20 \\ \text { (7) }(x-8)(x+2)=-24 & \text { (8) } \\ \text { (2x-3) }(x+1)=-2\end{array} \)

**(1)** Solve \( 4x^2 = 9x \). Rearrange the equation: \[ 4x^2 - 9x = 0 \] Factor out \( x \): \[ x(4x - 9) = 0 \] Set each factor equal to zero: \[ x = 0 \quad \text{or} \quad 4x - 9 = 0 \] Solving \( 4x - 9 = 0 \): \[ 4x = 9 \quad \Longrightarrow \quad x = \frac{9}{4} \] Thus, the solutions are: \[ x = 0 \quad \text{or} \quad x = \frac{9}{4} \] --- **(2)** Solve \( x^2 = 25 \). Take the square root of both sides: \[ x = \sqrt{25} \quad \text{or} \quad x = -\sqrt{25} \] Thus: \[ x = 5 \quad \text{or} \quad x = -5 \] --- **(3)** Solve \( x^2 - 5x = 14 \). Rearrange the equation: \[ x^2 - 5x - 14 = 0 \] Factor the quadratic. We look for two numbers that multiply to \(-14\) and add to \(-5\). These numbers are \(-7\) and \(2\): \[ (x - 7)(x + 2) = 0 \] Thus: \[ x = 7 \quad \text{or} \quad x = -2 \] --- **(4)** Solve \( -x^2 - 4x + 12 = 0 \). Multiply the entire equation by \(-1\) to simplify: \[ x^2 + 4x - 12 = 0 \] Factor the quadratic. We need two numbers that multiply to \(-12\) and add to \(4\); these are \(6\) and \(-2\): \[ (x + 6)(x - 2) = 0 \] Thus: \[ x = -6 \quad \text{or} \quad x = 2 \] --- **(5)** Solve \( 2x^2 - 5x + 3 = 0 \). Use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \(a = 2\), \(b = -5\), and \(c = 3\). Compute the discriminant: \[ b^2 - 4ac = (-5)^2 - 4(2)(3) = 25 - 24 = 1 \] Thus: \[ x = \frac{5 \pm \sqrt{1}}{4} = \frac{5 \pm 1}{4} \] So, \[ x = \frac{6}{4} = \frac{3}{2} \quad \text{or} \quad x = \frac{4}{4} = 1 \] --- **(6)** Solve \( 6x^2 + 7x = 20 \). Rearrange the equation: \[ 6x^2 + 7x - 20 = 0 \] Use the quadratic formula with \(a = 6\), \(b = 7\), and \(c = -20\): \[ x = \frac{-7 \pm \sqrt{7^2 - 4(6)(-20)}}{12} \] Calculate the discriminant: \[ 7^2 - 4(6)(-20) = 49 + 480 = 529 \] Since \( \sqrt{529} = 23 \), we have: \[ x = \frac{-7 \pm 23}{12} \] Thus, \[ x = \frac{-7 + 23}{12} = \frac{16}{12} = \frac{4}{3} \quad \text{or} \quad x = \frac{-7 - 23}{12} = \frac{-30}{12} = -\frac{5}{2} \] --- **(7)** Solve \( (x-8)(x+2) = -24 \). Expand the left-hand side: \[ x^2 - 8x + 2x - 16 = x^2 - 6x - 16 \] Thus, the equation is: \[ x^2 - 6x - 16 = -24 \] Add \( 24 \) to both sides: \[ x^2 - 6x + 8 = 0 \] Factor the quadratic: \[ (x - 2)(x - 4) = 0 \] Thus: \[ x = 2 \quad \text{or} \quad x = 4 \] --- **(8)** Solve \( (2x-3)(x+1) = -2 \). Expand the left-hand side: \[ 2x^2 + 2x - 3x - 3 = 2x^2 - x - 3 \] Thus, the equation becomes: \[ 2x^2 - x - 3 = -2 \] Add \(2\) to both sides: \[ 2x^2 - x - 1 = 0 \] Apply the quadratic formula with \(a = 2\), \(b = -1\), and \(c = -1\): \[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm \sqrt{9}}{4} \] Since \(\sqrt{9} = 3\), we have: \[ x = \frac{1 \pm 3}{4} \] Thus, \[ x = \frac{1+3}{4} = 1 \quad \text{or} \quad x = \frac{1-3}{4} = -\frac{1}{2} \]