4t2+2t-2=0

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Solution for 4t2+2t-2=0 equation: 



4t^2+2t-2=0

a = 4; b = 2; c = -2;
Δ = b2-4ac
Δ = 22-4·4·(-2)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6}{2*4}=\frac{-8}{8}
=-1
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6}{2*4}=\frac{4}{8}
=1/2
$

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