15=t2+4

Solution for 15=t2+4 equation:

15=t2+4

We move all terms to the left:

15-(t2+4)=0

We add all the numbers together, and all the variables

-(+t^2+4)+15=0

We get rid of parentheses

-t^2-4+15=0

We add all the numbers together, and all the variables

-1t^2+11=0

a = -1; b = 0; c = +11;
Δ = b2-4ac
Δ = 02-4·(-1)·11
Δ = 44
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:

$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{11}}{2*-1}=\frac{0-2\sqrt{11}}{-2}
=-\frac{2\sqrt{11}}{-2}
=-\frac{\sqrt{11}}{-1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{11}}{2*-1}=\frac{0+2\sqrt{11}}{-2}
=\frac{2\sqrt{11}}{-2}
=\frac{\sqrt{11}}{-1}
$