2t-t^2=-4

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

2t-t^2=-4

We move all terms to the left:

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

We add all the numbers together, and all the variables

-1t^2+2t+4=0

a = -1; b = 2; c = +4;
Δ = b2-4ac
Δ = 22-4·(-1)·4
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{5}}{2*-1}=\frac{-2-2\sqrt{5}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{5}}{2*-1}=\frac{-2+2\sqrt{5}}{-2}
$