3p^2=7p+20

Solution for 3p^2=7p+20 equation:

3p^2=7p+20

We move all terms to the left:

3p^2-(7p+20)=0

We get rid of parentheses

3p^2-7p-20=0

a = 3; b = -7; c = -20;
Δ = b2-4ac
Δ = -72-4·3·(-20)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-17}{2*3}=\frac{-10}{6}
=-1+2/3
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+17}{2*3}=\frac{24}{6}
=4
$