40t-5t2=40(t-2)-5(t-2)2

Solution for 40t-5t2=40(t-2)-5(t-2)2 equation:

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

We move all terms to the left:

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


We calculate terms in parentheses: -(40(t-2)-5(t-2)2), so:

40(t-2)-5(t-2)2

We multiply parentheses

40t-10t-80+20

We add all the numbers together, and all the variables

30t-60

Back to the equation:

-(30t-60)


We get rid of parentheses

-5t^2+40t-30t+60=0

We add all the numbers together, and all the variables

-5t^2+10t+60=0

a = -5; b = 10; c = +60;
Δ = b2-4ac
Δ = 102-4·(-5)·60
Δ = 1300
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:

$\sqrt{\Delta}=\sqrt{1300}=\sqrt{100*13}=\sqrt{100}*\sqrt{13}=10\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{13}}{2*-5}=\frac{-10-10\sqrt{13}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{13}}{2*-5}=\frac{-10+10\sqrt{13}}{-10}
$