4+t2=68

Solution for 4+t2=68 equation:

4+t2=68

We move all terms to the left:

4+t2-(68)=0

We add all the numbers together, and all the variables

t^2-64=0

a = 1; b = 0; c = -64;
Δ = b2-4ac
Δ = 02-4·1·(-64)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*1}=\frac{-16}{2}
=-8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*1}=\frac{16}{2}
=8
$