6p^2-29p+28=0

Solution for 6p^2-29p+28=0 equation:

6p^2-29p+28=0

a = 6; b = -29; c = +28;
Δ = b2-4ac
Δ = -292-4·6·28
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-13}{2*6}=\frac{16}{12}
=1+1/3
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+13}{2*6}=\frac{42}{12}
=3+1/2
$