5p^2=21p-4

Solution for 5p^2=21p-4 equation:

5p^2=21p-4

We move all terms to the left:

5p^2-(21p-4)=0

We get rid of parentheses

5p^2-21p+4=0

a = 5; b = -21; c = +4;
Δ = b2-4ac
Δ = -212-4·5·4
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-19}{2*5}=\frac{2}{10}
=1/5
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+19}{2*5}=\frac{40}{10}
=4
$