t^2+4t+1=3

Solution for t^2+4t+1=3 equation:

t^2+4t+1=3

We move all terms to the left:

t^2+4t+1-(3)=0

We add all the numbers together, and all the variables

t^2+4t-2=0

a = 1; b = 4; c = -2;
Δ = b2-4ac
Δ = 42-4·1·(-2)
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*1}=\frac{-4-2\sqrt{6}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{6}}{2*1}=\frac{-4+2\sqrt{6}}{2}
$