-3n2=-17n+20

Solution for -3n2=-17n+20 equation:

-3n^2=-17n+20

We move all terms to the left:

-3n^2-(-17n+20)=0

We get rid of parentheses

-3n^2+17n-20=0

a = -3; b = 17; c = -20;
Δ = b2-4ac
Δ = 172-4·(-3)·(-20)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-7}{2*-3}=\frac{-24}{-6}
=+4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+7}{2*-3}=\frac{-10}{-6}
=1+2/3
$