10. Solve the following equation: (round your answer/s to 1 dec.) (1 Point) \( x^{2}-7 x=9 \) No real solution \( x=-0,4 \) OR \( x=-2,3 \) \( x=-0,6 \) OR \( x=0,6 \) \( x=8,1 \) OR \( x=-1,1 \) \( x=1 \) OR \( x=0,4 \) 11. Solve the following equation: (round your answer/s to 1 dec.) (1 Point) \( x^{2}=3 x-9 \) \( x=-0,4 \) OR \( x=-2,3 \) \( x=-0,6 \) OR \( x=0,6 \) \( x=1 \) OR \( x=0,4 \) \( x=8,1 \) OR \( x=-1,1 \)

To solve the equation \(x^{2}-7x=9\), we can rearrange it to \(x^{2}-7x-9=0\). Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\), we identify \(a=1\), \(b=-7\), and \(c=-9\). Calculating gives us \(x = \frac{7 \pm \sqrt{(-7)^{2}-4(1)(-9)}}{2(1)}\), which simplifies to \(x = \frac{7 \pm \sqrt{49+36}}{2}\), or \(x = \frac{7 \pm \sqrt{85}}{2}\). This yields approximately \(x = 8.1\) or \(x = -1.1\). So the first equation gives us \(x = 8.1\) OR \(x = -1.1\). For the second equation \(x^{2}=3x-9\), we can rewrite it as \(x^{2}-3x+9=0\). Again applying the quadratic formula, we have \(a=1\), \(b=-3\), and \(c=9\). The discriminant \(b^{2}-4ac\) is calculated: \((-3)^{2}-4(1)(9) = 9 - 36 = -27\). Since the discriminant is negative, there are no real solutions for this equation. So, the solutions are: First Equation: \(x = 8.1\) OR \(x = -1.1\) Second Equation: No real solution Diving into the history, quadratic equations date back to ancient Babylonians, who managed to solve them geometrically—fascinating, right? They had a knack for solving problems with unknowns using methods that laid the groundwork for modern algebra! Now, when it comes to solving quadratics, a common mistake is neglecting to simplify the discriminant first! Always double-check your calculations for errors, especially when working with negative roots or potential imaginary solutions, so you don’t miss out on those real numbers hiding just beneath!