## Precalculus (6th Edition) Blitzer

$15.6$ years old
$f(x)=-2.32x^2+76.58x-559.87$ and $12 \leq x \leq 17$ ; $f(x)=70$ $-2.32x^2+76.58x-559.87=70$ or, $2.32x^2-76.58x+629.87=0$ $x=\dfrac{-(-76.58) \pm \sqrt {(76.58)^2-4 \times 2.32 \times 629.87}}{(2)(2.32)}$ or, $x=\dfrac{76.58 \pm 4.3935}{4.64}$ So, $x_1 \approx 17.5; x_2 \approx 15.6$ Because $12 \leq x \leq 17$, only the value of $x_2$ applies. So, $x_2 \approx 15.6$