f^2-13f+12=0

Solution for f^2-13f+12=0 equation:

f^2-13f+12=0

a = 1; b = -13; c = +12;
Δ = b2-4ac
Δ = -132-4·1·12
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-11}{2*1}=\frac{2}{2}
=1
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+11}{2*1}=\frac{24}{2}
=12
$