9t^2-28t+3=0

Solution for 9t^2-28t+3=0 equation:

9t^2-28t+3=0

a = 9; b = -28; c = +3;
Δ = b2-4ac
Δ = -282-4·9·3
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-26}{2*9}=\frac{2}{18}
=1/9
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+26}{2*9}=\frac{54}{18}
=3
$