t+2/4=t2

Solution for t+2/4=t2 equation:

t+2/4=t2

We move all terms to the left:

t+2/4-(t2)=0

We add all the numbers together, and all the variables

-1t^2+t+2/4=0

We multiply all the terms by the denominator


-1t^2*4+t*4+2=0

Wy multiply elements

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

a = -4; b = 4; c = +2;
Δ = b2-4ac
Δ = 42-4·(-4)·2
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{3}}{2*-4}=\frac{-4-4\sqrt{3}}{-8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{3}}{2*-4}=\frac{-4+4\sqrt{3}}{-8}
$