t2+3=4

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Solution for t2+3=4 equation:



t2+3=4
We move all terms to the left:
t2+3-(4)=0
We add all the numbers together, and all the variables
t^2-1=0
a = 1; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·1·(-1)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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

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