2t^2-15t+18=0

Solution for 2t^2-15t+18=0 equation:

2t^2-15t+18=0

a = 2; b = -15; c = +18;
Δ = b2-4ac
Δ = -152-4·2·18
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-9}{2*2}=\frac{6}{4}
=1+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+9}{2*2}=\frac{24}{4}
=6
$