t^2-6T=18

Solution for t^2-6T=18 equation:

t^2-6=18

We move all terms to the left:

t^2-6-(18)=0

We add all the numbers together, and all the variables

t^2-24=0

a = 1; b = 0; c = -24;
Δ = b2-4ac
Δ = 02-4·1·(-24)
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*1}=\frac{0-4\sqrt{6}}{2}
=-\frac{4\sqrt{6}}{2}
=-2\sqrt{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*1}=\frac{0+4\sqrt{6}}{2}
=\frac{4\sqrt{6}}{2}
=2\sqrt{6}
$