36=28+t2

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Solution for 36=28+t2 equation:



36=28+t2
We move all terms to the left:
36-(28+t2)=0
We add all the numbers together, and all the variables
-(+t^2+28)+36=0
We get rid of parentheses
-t^2-28+36=0
We add all the numbers together, and all the variables
-1t^2+8=0
a = -1; b = 0; c = +8;
Δ = b2-4ac
Δ = 02-4·(-1)·8
Δ = 32
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:
$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*-1}=\frac{0-4\sqrt{2}}{-2} =-\frac{4\sqrt{2}}{-2} =-\frac{2\sqrt{2}}{-1} $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*-1}=\frac{0+4\sqrt{2}}{-2} =\frac{4\sqrt{2}}{-2} =\frac{2\sqrt{2}}{-1} $

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