t=0.1(400-40t+t2)

Solution for t=0.1(400-40t+t2) equation:

t=0.1(400-40t+t2)

We move all terms to the left:

t-(0.1(400-40t+t2))=0

We add all the numbers together, and all the variables

-(0.1(-40t+t^2+400))+t=0


We calculate terms in parentheses: -(0.1(-40t+t^2+400)), so:

0.1(-40t+t^2+400)

We multiply parentheses

0.1t^2-4t+40

Back to the equation:

-(0.1t^2-4t+40)


We add all the numbers together, and all the variables

t-(0.1t^2-4t+40)=0

We get rid of parentheses

-0.1t^2+t+4t-40=0

We add all the numbers together, and all the variables

-0.1t^2+5t-40=0

a = -0.1; b = 5; c = -40;
Δ = b2-4ac
Δ = 52-4·(-0.1)·(-40)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3}{2*-0.1}=\frac{-8}{-0.2}
=+40
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3}{2*-0.1}=\frac{-2}{-0.2}
=+10
$