0=-16t^2+80+200

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Solution for 0=-16t^2+80+200 equation:



0=-16t^2+80+200
We move all terms to the left:
0-(-16t^2+80+200)=0
We add all the numbers together, and all the variables
-(-16t^2+80+200)=0
We get rid of parentheses
16t^2-80-200=0
We add all the numbers together, and all the variables
16t^2-280=0
a = 16; b = 0; c = -280;
Δ = b2-4ac
Δ = 02-4·16·(-280)
Δ = 17920
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:
$\sqrt{\Delta}=\sqrt{17920}=\sqrt{256*70}=\sqrt{256}*\sqrt{70}=16\sqrt{70}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{70}}{2*16}=\frac{0-16\sqrt{70}}{32} =-\frac{16\sqrt{70}}{32} =-\frac{\sqrt{70}}{2} $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{70}}{2*16}=\frac{0+16\sqrt{70}}{32} =\frac{16\sqrt{70}}{32} =\frac{\sqrt{70}}{2} $

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