t^2+9=13t

Solution for t^2+9=13t equation:

t^2+9=13t

We move all terms to the left:

t^2+9-(13t)=0

a = 1; b = -13; c = +9;
Δ = b2-4ac
Δ = -132-4·1·9
Δ = 133
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-\sqrt{133}}{2*1}=\frac{13-\sqrt{133}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{133}}{2*1}=\frac{13+\sqrt{133}}{2}
$