4t^2-16t-65=0

Solution for 4t^2-16t-65=0 equation:

4t^2-16t-65=0

a = 4; b = -16; c = -65;
Δ = b2-4ac
Δ = -162-4·4·(-65)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-36}{2*4}=\frac{-20}{8}
=-2+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+36}{2*4}=\frac{52}{8}
=6+1/2
$