4t+16=t2-2t

Solution for 4t+16=t2-2t equation:

4t+16=t2-2t

We move all terms to the left:

4t+16-(t2-2t)=0

We add all the numbers together, and all the variables

-(+t^2-2t)+4t+16=0

We get rid of parentheses

-t^2+2t+4t+16=0

We add all the numbers together, and all the variables

-1t^2+6t+16=0

a = -1; b = 6; c = +16;
Δ = b2-4ac
Δ = 62-4·(-1)·16
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-10}{2*-1}=\frac{-16}{-2}
=+8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+10}{2*-1}=\frac{4}{-2}
=-2
$