t^2-11t+18=0t=

Solution for t^2-11t+18=0t= equation:

t^2-11t+18=0t=

We move all terms to the left:

t^2-11t+18-(0t)=0

We add all the numbers together, and all the variables

t^2-12t+18=0

a = 1; b = -12; c = +18;
Δ = b2-4ac
Δ = -122-4·1·18
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{2}}{2*1}=\frac{12-6\sqrt{2}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{2}}{2*1}=\frac{12+6\sqrt{2}}{2}
$