-3t^2+13t=2

Solution for -3t^2+13t=2 equation:

-3t^2+13t=2

We move all terms to the left:

-3t^2+13t-(2)=0

a = -3; b = 13; c = -2;
Δ = b2-4ac
Δ = 132-4·(-3)·(-2)
Δ = 145
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{145}}{2*-3}=\frac{-13-\sqrt{145}}{-6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{145}}{2*-3}=\frac{-13+\sqrt{145}}{-6}
$