4n^2=-n+95

Solution for 4n^2=-n+95 equation:

4n^2=-n+95

We move all terms to the left:

4n^2-(-n+95)=0

We add all the numbers together, and all the variables

4n^2-(-1n+95)=0

We get rid of parentheses

4n^2+1n-95=0

We add all the numbers together, and all the variables

4n^2+n-95=0

a = 4; b = 1; c = -95;
Δ = b2-4ac
Δ = 12-4·4·(-95)
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-39}{2*4}=\frac{-40}{8}
=-5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+39}{2*4}=\frac{38}{8}
=4+3/4
$