((t^2)-25)/5t=0

Solution for ((t^2)-25)/5t=0 equation:

((t^2)-25)/5t=0


Domain of the equation: 5t!=0

t!=0/5

t!=0

t∈R


We multiply all the terms by the denominator


(t^2-25)=0

We get rid of parentheses

t^2-25=0

a = 1; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·1·(-25)
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{-(0)-10}{2*1}=\frac{-10}{2}
=-5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10}{2*1}=\frac{10}{2}
=5
$