T2+3=t2

Solution for T2+3=t2 equation:

2+3=T2

We move all terms to the left:

2+3-(T2)=0

We add all the numbers together, and all the variables

-1T^2+5=0

a = -1; b = 0; c = +5;
Δ = b2-4ac
Δ = 02-4·(-1)·5
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{5}}{2*-1}=\frac{0-2\sqrt{5}}{-2}
=-\frac{2\sqrt{5}}{-2}
=-\frac{\sqrt{5}}{-1}
$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{5}}{2*-1}=\frac{0+2\sqrt{5}}{-2}
=\frac{2\sqrt{5}}{-2}
=\frac{\sqrt{5}}{-1}
$