t^2-4+3=0

Solution for t^2-4+3=0 equation:

t^2-4+3=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
$