(80-40t)/t^2=5

Solution for (80-40t)/t^2=5 equation:

(80-40t)/t^2=5

We move all terms to the left:

(80-40t)/t^2-(5)=0


Domain of the equation: t^2!=0

t^2!=0/

t^2!=√0

t!=0

t∈R


We add all the numbers together, and all the variables

(-40t+80)/t^2-5=0

We multiply all the terms by the denominator


(-40t+80)-5*t^2=0

We add all the numbers together, and all the variables

-5t^2+(-40t+80)=0

We get rid of parentheses

-5t^2-40t+80=0

a = -5; b = -40; c = +80;
Δ = b2-4ac
Δ = -402-4·(-5)·80
Δ = 3200
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:

$\sqrt{\Delta}=\sqrt{3200}=\sqrt{1600*2}=\sqrt{1600}*\sqrt{2}=40\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-40\sqrt{2}}{2*-5}=\frac{40-40\sqrt{2}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+40\sqrt{2}}{2*-5}=\frac{40+40\sqrt{2}}{-10}
$