12.5=t^2-4t+4

Solution for 12.5=t^2-4t+4 equation:

12.5=t^2-4t+4

We move all terms to the left:

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

We get rid of parentheses

-t^2+4t-4+12.5=0

We add all the numbers together, and all the variables

-1t^2+4t+8.5=0

a = -1; b = 4; c = +8.5;
Δ = b2-4ac
Δ = 42-4·(-1)·8.5
Δ = 50
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:

$\sqrt{\Delta}=\sqrt{50}=\sqrt{25*2}=\sqrt{25}*\sqrt{2}=5\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-5\sqrt{2}}{2*-1}=\frac{-4-5\sqrt{2}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+5\sqrt{2}}{2*-1}=\frac{-4+5\sqrt{2}}{-2}
$