## Calculus Concepts: An Informal Approach to the Mathematics of Change 5th Edition

If $t(x)=10$, then $t=1.6$ or $t=-1$ If $t(x)=15$, then $t\approx1.940$ or $t\approx-1.340$
$t(x)=5x^{2}-3x+2$ This problem is much harder, because it requires us to work backwards for the $x$ values. What you must do is solve the equation as $5x^{2}-3x+2=(10,15)$ So to set up, we will do $5x^{2}-3x+2=10$ and solve. $5x^{2}-3x+2=10$ $5x^{2}-3x-8=0$ (Subtract 10 from both sides) Use the Quadratic Formula to solve both roots:$(\frac{-b(+/-)\sqrt {b^{2}-4ac}}{2a})$ Positive Root: $\frac{(-(-3)+\sqrt {3^{2}-4(5)(-8)}}{2(5)}$ (Substitute) $\frac{3+\sqrt {169}}{10}=1.6$ (Simplify and solve) Negative Root: $\frac{(-(-3)-\sqrt {3^{2}-4(5)(-8)}}{2(5)}$ (Substitute) $\frac{3-\sqrt {169}}{10}=-1$ (Simplify and solve) You can solve the other part of the problem in the same fashion. $5x^{2}-3x+2=15$ $5x^{2}-3x-13=0$ Positive Root: $\frac{(-(-3)+\sqrt {3^{2}-4(5)(-13)}}{2(5)}$ $\frac{3+\sqrt {269}}{10}\approx1.940$ Negative Root: $\frac{(-(-3)-\sqrt {3^{2}-4(5)(-13)}}{2(5)}$ $\frac{3-\sqrt {269}}{10}\approx$-1.340