0.5t^2+2t-6=0

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Solution for 0.5t^2+2t-6=0 equation: 



0.5t^2+2t-6=0

a = 0.5; b = 2; c = -6;
Δ = b2-4ac
Δ = 22-4·0.5·(-6)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4}{2*0.5}=\frac{-6}{1}
=-6
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4}{2*0.5}=\frac{2}{1}
=2
$

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