5t^2-20=0

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Solution for 5t^2-20=0 equation: 



5t^2-20=0

a = 5; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·5·(-20)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*5}=\frac{-20}{10}
=-2
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*5}=\frac{20}{10}
=2
$

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