t^2-10t+12=0

Solution for t^2-10t+12=0 equation:

t^2-10t+12=0

a = 1; b = -10; c = +12;
Δ = b2-4ac
Δ = -102-4·1·12
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{13}}{2*1}=\frac{10-2\sqrt{13}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{13}}{2*1}=\frac{10+2\sqrt{13}}{2}
$