2t^2-98=0

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Solution for 2t^2-98=0 equation: 



2t^2-98=0

a = 2; b = 0; c = -98;
Δ = b2-4ac
Δ = 02-4·2·(-98)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-28}{2*2}=\frac{-28}{4}
=-7
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+28}{2*2}=\frac{28}{4}
=7
$

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