0=t(10-4.905t)

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Solution for 0=t(10-4.905t) equation:



0=t(10-4.905t)
We move all terms to the left:
0-(t(10-4.905t))=0
We add all the numbers together, and all the variables
-(t(-4.905t+10))+0=0
We add all the numbers together, and all the variables
-(t(-4.905t+10))=0
We calculate terms in parentheses: -(t(-4.905t+10)), so:
t(-4.905t+10)
We multiply parentheses
-4t^2+10t
Back to the equation:
-(-4t^2+10t)
We get rid of parentheses
4t^2-10t=0
a = 4; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·4·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10}{2*4}=\frac{0}{8} =0 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*4}=\frac{20}{8} =2+1/2 $

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