1. **State the problem:** We need to find the input values $x$ such that $t(x) = 12$ and $t(x) = 17$ for the function $$t(x) = 5x^2 - 3x + 4.$$ 2. **Use the formula:** Set the function equal to the given output values and solve for $x$. 3. **Solve for $t(x) = 12$: ** $$5x^2 - 3x + 4 = 12$$ Subtract 12 from both sides: $$5x^2 - 3x + 4 - 12 = 0$$ $$5x^2 - 3x - 8 = 0$$ 4. **Apply the quadratic formula:** For $ax^2 + bx + c = 0$, $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=5$, $b=-3$, $c=-8$. Calculate the discriminant: $$\Delta = (-3)^2 - 4(5)(-8) = 9 + 160 = 169$$ Calculate the roots: $$x = \frac{-(-3) \pm \sqrt{169}}{2 \times 5} = \frac{3 \pm 13}{10}$$ So, $$x_1 = \frac{3 + 13}{10} = \frac{16}{10} = 1.6$$ $$x_2 = \frac{3 - 13}{10} = \frac{-10}{10} = -1$$ 5. **Solve for $t(x) = 17$: ** $$5x^2 - 3x + 4 = 17$$ Subtract 17 from both sides: $$5x^2 - 3x + 4 - 17 = 0$$ $$5x^2 - 3x - 13 = 0$$ 6. **Apply the quadratic formula again:** Here, $a=5$, $b=-3$, $c=-13$. Calculate the discriminant: $$\Delta = (-3)^2 - 4(5)(-13) = 9 + 260 = 269$$ Calculate the roots: $$x = \frac{-(-3) \pm \sqrt{269}}{2 \times 5} = \frac{3 \pm \sqrt{269}}{10}$$ Approximate $\sqrt{269} \approx 16.401$: $$x_1 = \frac{3 + 16.401}{10} = \frac{19.401}{10} = 1.940$$ $$x_2 = \frac{3 - 16.401}{10} = \frac{-13.401}{10} = -1.340$$ **Final answers:** - For $t(x) = 12$, $x = 1.6, -1$ - For $t(x) = 17$, $x = 1.940, -1.340$