t^2+26.5t+169/4=0

Solution for t^2+26.5t+169/4=0 equation:

t^2+26.5t+169/4=0

We multiply all the terms by the denominator


t^2*4+(26.5t)*4+169=0

We add all the numbers together, and all the variables

t^2*4+(+26.5t)*4+169=0

We multiply parentheses

t^2*4+104t+169=0

Wy multiply elements

4t^2+104t+169=0

a = 4; b = 104; c = +169;
Δ = b2-4ac
Δ = 1042-4·4·169
Δ = 8112
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:

$\sqrt{\Delta}=\sqrt{8112}=\sqrt{2704*3}=\sqrt{2704}*\sqrt{3}=52\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(104)-52\sqrt{3}}{2*4}=\frac{-104-52\sqrt{3}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(104)+52\sqrt{3}}{2*4}=\frac{-104+52\sqrt{3}}{8}
$