-2t^2+7t+39=0

Solution for -2t^2+7t+39=0 equation:

-2t^2+7t+39=0

a = -2; b = 7; c = +39;
Δ = b2-4ac
Δ = 72-4·(-2)·39
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-19}{2*-2}=\frac{-26}{-4}
=6+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+19}{2*-2}=\frac{12}{-4}
=-3
$