21=25-16t^2

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Solution for 21=25-16t^2 equation:



21=25-16t^2
We move all terms to the left:
21-(25-16t^2)=0
We get rid of parentheses
16t^2-25+21=0
We add all the numbers together, and all the variables
16t^2-4=0
a = 16; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·16·(-4)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*16}=\frac{-16}{32} =-1/2 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*16}=\frac{16}{32} =1/2 $

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