Exercises \( \begin{array}{ll}\text { What are the solutions for each equation? Use the Quadratic Formula. } \\ \begin{array}{ll}\text { 1. }-x^{2}+7 x-3=0 & \text { 2. } x^{2}+6 x=10 \\ \text { 3. } 2 x^{2}=4 x+3 & \text { 4. } 4 x^{2}+81=36 x\end{array} \\ \text { Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. }\end{array} \)

To solve the equations using the Quadratic Formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): 1. For \(-x^2 + 7x - 3 = 0\): Rearranging gives \(x^2 - 7x + 3 = 0\). Here, \(a = 1\), \(b = -7\), and \(c = 3\). Plugging into the formula results in: \( x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} = \frac{7 \pm \sqrt{49 - 12}}{2} = \frac{7 \pm \sqrt{37}}{2} \). Thus, the solutions are \( x = \frac{7 + \sqrt{37}}{2} \) and \( x = \frac{7 - \sqrt{37}}{2} \). 2. For \(x^2 + 6x = 10\): Rearranging gives \(x^2 + 6x - 10 = 0\). Here, \(a = 1\), \(b = 6\), and \(c = -10\). Plugging into the formula results in: \( x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} = \frac{-6 \pm \sqrt{36 + 40}}{2} = \frac{-6 \pm \sqrt{76}}{2} = \frac{-6 \pm 2\sqrt{19}}{2} = -3 \pm \sqrt{19} \). Thus, the solutions are \( x = -3 + \sqrt{19} \) and \( x = -3 - \sqrt{19} \). 3. For \(2x^2 = 4x + 3\): Rearranging gives \(2x^2 - 4x - 3 = 0\). Here, \(a = 2\), \(b = -4\), and \(c = -3\). Plugging into the formula results in: \( x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} = \frac{4 \pm \sqrt{16 + 24}}{4} = \frac{4 \pm \sqrt{40}}{4} = \frac{4 \pm 2\sqrt{10}}{4} = 1 \pm \frac{\sqrt{10}}{2} \). Thus, the solutions are \( x = 1 + \frac{\sqrt{10}}{2} \) and \( x = 1 - \frac{\sqrt{10}}{2} \). 4. For \(4x^2 + 81 = 36x\): Rearranging gives \(4x^2 - 36x + 81 = 0\). Here, \(a = 4\), \(b = -36\), and \(c = 81\). Plugging into the formula results in: \( x = \frac{36 \pm \sqrt{(-36)^2 - 4 \cdot 4 \cdot 81}}{2 \cdot 4} = \frac{36 \pm \sqrt{1296 - 1296}}{8} = \frac{36 \pm 0}{8} = \frac{36}{8} = \frac{9}{2} \). Thus, the solution is \( x = \frac{9}{2} \) (a double root). Happy solving!