3n^2=-7n+20

Solution for 3n^2=-7n+20 equation:

3n^2=-7n+20

We move all terms to the left:

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

We get rid of parentheses

3n^2+7n-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:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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