Answer:
[tex]t=5\ sec[/tex]
Step-by-step explanation:
we have
[tex]-t^{2} +4t+5=0[/tex]
Solve the quadratic equation to find the zero's or x-intercepts
Remember that
The x-intercepts are the values of x when the value of y is equal to zero ( water balloon reach the ground)
we know that
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]-t^{2} +4t+5=0[/tex]
so
[tex]a=-1\\b=4\\c=5[/tex]
substitute in the formula
[tex]t=\frac{-4(+/-)\sqrt{4^{2}-4(-1)(5)}} {2(-1)}[/tex]
[tex]t=\frac{-4(+/-)\sqrt{36}} {-2}[/tex]
[tex]t=\frac{-4(+/-)6} {-2}[/tex]
[tex]t=\frac{-4(+)6} {-2}=-1[/tex]
[tex]t=\frac{-4(-)6} {-2}=5[/tex]
therefore
the solution is [tex]t=5\ sec[/tex]
