t=-t^2+4t+9

Solution for t=-t^2+4t+9 equation:

t=-t^2+4t+9

We move all terms to the left:

t-(-t^2+4t+9)=0

We get rid of parentheses

t^2-4t+t-9=0

We add all the numbers together, and all the variables

t^2-3t-9=0

a = 1; b = -3; c = -9;
Δ = b2-4ac
Δ = -32-4·1·(-9)
Δ = 45
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{5}}{2*1}=\frac{3-3\sqrt{5}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{5}}{2*1}=\frac{3+3\sqrt{5}}{2}
$