1. **Solve the quadratic equation:** $12x^2 - 40 = 17x$ to 2 decimal places. Step 1: Rewrite the equation in standard form: $$12x^2 - 17x - 40 = 0$$ Step 2: Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=12$, $b=-17$, and $c=-40$. Step 3: Calculate the discriminant: $$\Delta = (-17)^2 - 4 \times 12 \times (-40) = 289 + 1920 = 2209$$ Step 4: Find the square root of the discriminant: $$\sqrt{2209} = 47$$ Step 5: Calculate the two solutions: $$x_1 = \frac{17 + 47}{24} = \frac{64}{24} = 2.67$$ $$x_2 = \frac{17 - 47}{24} = \frac{-30}{24} = -1.25$$ Answer: $x = 2.67$ or $x = -1.25$. 2. **Find the points of intersection algebraically for the system:** $$y = -x^2 + 6x - 5$$ $$y = -4x + 19$$ Step 1: Set the two expressions for $y$ equal to each other: $$-x^2 + 6x - 5 = -4x + 19$$ Step 2: Rearrange to form a quadratic equation: $$-x^2 + 6x - 5 + 4x - 19 = 0$$ $$-x^2 + 10x - 24 = 0$$ Multiply both sides by $-1$ for simplicity: $$x^2 - 10x + 24 = 0$$ Step 3: Factor the quadratic: $$(x - 6)(x - 4) = 0$$ Step 4: Solve for $x$: $$x = 6 \quad \text{or} \quad x = 4$$ Step 5: Substitute back into $y = -4x + 19$ to find $y$ values: For $x=6$: $$y = -4(6) + 19 = -24 + 19 = -5$$ For $x=4$: $$y = -4(4) + 19 = -16 + 19 = 3$$ Answer: Points of intersection are $(6, -5)$ and $(4, 3)$.