252/t(2)+18=25

Solution for 252/t(2)+18=25 equation:

252/t2+18=25

We move all terms to the left:

252/t2+18-(25)=0


Domain of the equation: t2!=0

t^2!=0/

t^2!=√0

t!=0

t∈R


We add all the numbers together, and all the variables

252/t2-7=0

We multiply all the terms by the denominator


-7*t2+252=0

We add all the numbers together, and all the variables

-7t^2+252=0

a = -7; b = 0; c = +252;
Δ = b2-4ac
Δ = 02-4·(-7)·252
Δ = 7056
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{7056}=84$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-84}{2*-7}=\frac{-84}{-14}
=+6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+84}{2*-7}=\frac{84}{-14}
=-6
$