-4t^2+2t+2=0

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Solution for -4t^2+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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