t=t^2+6t-27

Solution for t=t^2+6t-27 equation:

t=t^2+6t-27

We move all terms to the left:

t-(t^2+6t-27)=0

We get rid of parentheses

-t^2+t-6t+27=0

We add all the numbers together, and all the variables

-1t^2-5t+27=0

a = -1; b = -5; c = +27;
Δ = b2-4ac
Δ = -52-4·(-1)·27
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{-(-5)-\sqrt{133}}{2*-1}=\frac{5-\sqrt{133}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{133}}{2*-1}=\frac{5+\sqrt{133}}{-2}
$