-5+t^2=14

Solution for -5+t^2=14 equation:

-5+t^2=14

We move all terms to the left:

-5+t^2-(14)=0

We add all the numbers together, and all the variables

t^2-19=0

a = 1; b = 0; c = -19;
Δ = b2-4ac
Δ = 02-4·1·(-19)
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{19}}{2*1}=\frac{0-2\sqrt{19}}{2}
=-\frac{2\sqrt{19}}{2}
=-\sqrt{19}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{19}}{2*1}=\frac{0+2\sqrt{19}}{2}
=\frac{2\sqrt{19}}{2}
=\sqrt{19}
$