1-2t^2+t=0

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Solution for 1-2t^2+t=0 equation: 



1-2t^2+t=0

a = -2; b = 1; c = +1;
Δ = b2-4ac
Δ = 12-4·(-2)·1
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$


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


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

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