0=-16t^2+0t+25

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Solution for 0=-16t^2+0t+25 equation:



0=-16t^2+0t+25
We move all terms to the left:
0-(-16t^2+0t+25)=0
We add all the numbers together, and all the variables
-(-16t^2+0t+25)=0
We get rid of parentheses
16t^2-0t-25=0
We add all the numbers together, and all the variables
16t^2-1t-25=0
a = 16; b = -1; c = -25;
Δ = b2-4ac
Δ = -12-4·16·(-25)
Δ = 1601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{1601}}{2*16}=\frac{1-\sqrt{1601}}{32} $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{1601}}{2*16}=\frac{1+\sqrt{1601}}{32} $

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