t^2+2t-1/2=0

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

t^2+2t-1/2=0

We multiply all the terms by the denominator


t^2*2+2t*2-1=0

Wy multiply elements

2t^2+4t-1=0

a = 2; b = 4; c = -1;
Δ = b2-4ac
Δ = 42-4·2·(-1)
Δ = 24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{6}}{2*2}=\frac{-4-2\sqrt{6}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{6}}{2*2}=\frac{-4+2\sqrt{6}}{4}
$