2t2-18=0

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Solution for 2t2-18=0 equation: 



2t^2-18=0

a = 2; b = 0; c = -18;
Δ = b2-4ac
Δ = 02-4·2·(-18)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*2}=\frac{-12}{4}
=-3
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*2}=\frac{12}{4}
=3
$

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