7=-5n(n-4)-3

Solution for 7=-5n(n-4)-3 equation:

7=-5n(n-4)-3

We move all terms to the left:

7-(-5n(n-4)-3)=0


We calculate terms in parentheses: -(-5n(n-4)-3), so:

-5n(n-4)-3

We multiply parentheses

-5n^2+20n-3

Back to the equation:

-(-5n^2+20n-3)


We get rid of parentheses

5n^2-20n+3+7=0

We add all the numbers together, and all the variables

5n^2-20n+10=0

a = 5; b = -20; c = +10;
Δ = b2-4ac
Δ = -202-4·5·10
Δ = 200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-10\sqrt{2}}{2*5}=\frac{20-10\sqrt{2}}{10}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+10\sqrt{2}}{2*5}=\frac{20+10\sqrt{2}}{10}
$