t^2-18t+18=121

Solution for t^2-18t+18=121 equation:

t^2-18t+18=121

We move all terms to the left:

t^2-18t+18-(121)=0

We add all the numbers together, and all the variables

t^2-18t-103=0

a = 1; b = -18; c = -103;
Δ = b2-4ac
Δ = -182-4·1·(-103)
Δ = 736
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:

$\sqrt{\Delta}=\sqrt{736}=\sqrt{16*46}=\sqrt{16}*\sqrt{46}=4\sqrt{46}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-4\sqrt{46}}{2*1}=\frac{18-4\sqrt{46}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+4\sqrt{46}}{2*1}=\frac{18+4\sqrt{46}}{2}
$