t2+16t=2t+32

Solution for t2+16t=2t+32 equation:

t2+16t=2t+32

We move all terms to the left:

t2+16t-(2t+32)=0

We add all the numbers together, and all the variables

t^2+16t-(2t+32)=0

We get rid of parentheses

t^2+16t-2t-32=0

We add all the numbers together, and all the variables

t^2+14t-32=0

a = 1; b = 14; c = -32;
Δ = b2-4ac
Δ = 142-4·1·(-32)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-18}{2*1}=\frac{-32}{2}
=-16
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+18}{2*1}=\frac{4}{2}
=2
$