11=6+t2

Solution for 11=6+t2 equation:

11=6+t2

We move all terms to the left:

11-(6+t2)=0

We add all the numbers together, and all the variables

-(+t^2+6)+11=0

We get rid of parentheses

-t^2-6+11=0

We add all the numbers together, and all the variables

-1t^2+5=0

a = -1; b = 0; c = +5;
Δ = b2-4ac
Δ = 02-4·(-1)·5
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{5}}{2*-1}=\frac{0-2\sqrt{5}}{-2}
=-\frac{2\sqrt{5}}{-2}
=-\frac{\sqrt{5}}{-1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{5}}{2*-1}=\frac{0+2\sqrt{5}}{-2}
=\frac{2\sqrt{5}}{-2}
=\frac{\sqrt{5}}{-1}
$