t^2-t+6=18

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

t^2-t+6=18

We move all terms to the left:

t^2-t+6-(18)=0

We add all the numbers together, and all the variables

t^2-1t-12=0

a = 1; b = -1; c = -12;
Δ = b2-4ac
Δ = -12-4·1·(-12)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-7}{2*1}=\frac{-6}{2}
=-3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+7}{2*1}=\frac{8}{2}
=4
$