20p^2=14p

Solution for 20p^2=14p equation:

20p^2=14p

We move all terms to the left:

20p^2-(14p)=0

a = 20; b = -14; c = 0;
Δ = b2-4ac
Δ = -142-4·20·0
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-14}{2*20}=\frac{0}{40}
=0
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+14}{2*20}=\frac{28}{40}
=7/10
$