12t^2-56t+65=0

Solution for 12t^2-56t+65=0 equation:

12t^2-56t+65=0

a = 12; b = -56; c = +65;
Δ = b2-4ac
Δ = -562-4·12·65
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-56)-4}{2*12}=\frac{52}{24}
=2+1/6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-56)+4}{2*12}=\frac{60}{24}
=2+1/2
$