-5t^2+12t+65=0

Solution for -5t^2+12t+65=0 equation:

-5t^2+12t+65=0

a = -5; b = 12; c = +65;
Δ = b2-4ac
Δ = 122-4·(-5)·65
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-38}{2*-5}=\frac{-50}{-10}
=+5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+38}{2*-5}=\frac{26}{-10}
=-2+3/5
$