0.51=t+t^2

Solution for 0.51=t+t^2 equation:

0.51=t+t^2

We move all terms to the left:

0.51-(t+t^2)=0

We get rid of parentheses

-t^2-t+0.51=0

We add all the numbers together, and all the variables

-1t^2-1t+0.51=0

a = -1; b = -1; c = +0.51;
Δ = b2-4ac
Δ = -12-4·(-1)·0.51
Δ = 3.04
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{3.04}}{2*-1}=\frac{1-\sqrt{3.04}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{3.04}}{2*-1}=\frac{1+\sqrt{3.04}}{-2}
$