t^2-12t+5=0

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

t^2-12t+5=0

a = 1; b = -12; c = +5;
Δ = b2-4ac
Δ = -122-4·1·5
Δ = 124
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{124}=\sqrt{4*31}=\sqrt{4}*\sqrt{31}=2\sqrt{31}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{31}}{2*1}=\frac{12-2\sqrt{31}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{31}}{2*1}=\frac{12+2\sqrt{31}}{2}
$