0=50000(2t-1)(t-5)

Solution for 0=50000(2t-1)(t-5) equation:

0=50000(2t-1)(t-5)

We move all terms to the left:

0-(50000(2t-1)(t-5))=0

We add all the numbers together, and all the variables

-(50000(2t-1)(t-5))=0

We multiply parentheses ..

-(50000(+2t^2-10t-1t+5))=0


We calculate terms in parentheses: -(50000(+2t^2-10t-1t+5)), so:

50000(+2t^2-10t-1t+5)

We multiply parentheses

100000t^2-500000t-50000t+250000

We add all the numbers together, and all the variables

100000t^2-550000t+250000

Back to the equation:

-(100000t^2-550000t+250000)


We get rid of parentheses

-100000t^2+550000t-250000=0

a = -100000; b = 550000; c = -250000;
Δ = b2-4ac
Δ = 5500002-4·(-100000)·(-250000)
Δ = 202500000000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{202500000000}=450000$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(550000)-450000}{2*-100000}=\frac{-1000000}{-200000}
=+5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(550000)+450000}{2*-100000}=\frac{-100000}{-200000}
=1/2
$