2t^2-19t+9=0

Solution for 2t^2-19t+9=0 equation:

2t^2-19t+9=0

a = 2; b = -19; c = +9;
Δ = b2-4ac
Δ = -192-4·2·9
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-17}{2*2}=\frac{2}{4}
=1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+17}{2*2}=\frac{36}{4}
=9
$