8t^2-65t+132=0

Solution for 8t^2-65t+132=0 equation:

8t^2-65t+132=0

a = 8; b = -65; c = +132;
Δ = b2-4ac
Δ = -652-4·8·132
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-65)-1}{2*8}=\frac{64}{16}
=4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-65)+1}{2*8}=\frac{66}{16}
=4+1/8
$