(5t^2-16t+3)/t=0

Solution for (5t^2-16t+3)/t=0 equation:

(5t^2-16t+3)/t=0


Domain of the equation: t!=0

t∈R


We multiply all the terms by the denominator


(5t^2-16t+3)=0

We get rid of parentheses

5t^2-16t+3=0

a = 5; b = -16; c = +3;
Δ = b2-4ac
Δ = -162-4·5·3
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-14}{2*5}=\frac{2}{10}
=1/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+14}{2*5}=\frac{30}{10}
=3
$