V(t)=-5t^2+16t+9

Solution for V(t)=-5t^2+16t+9 equation:

(V)=-5V^2+16V+9

We move all terms to the left:

(V)-(-5V^2+16V+9)=0

We get rid of parentheses

5V^2-16V+V-9=0

We add all the numbers together, and all the variables

5V^2-15V-9=0

a = 5; b = -15; c = -9;
Δ = b2-4ac
Δ = -152-4·5·(-9)
Δ = 405
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{405}=\sqrt{81*5}=\sqrt{81}*\sqrt{5}=9\sqrt{5}$
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-9\sqrt{5}}{2*5}=\frac{15-9\sqrt{5}}{10}
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+9\sqrt{5}}{2*5}=\frac{15+9\sqrt{5}}{10}
$