n^2+4n-7=28

Solution for n^2+4n-7=28 equation:

n^2+4n-7=28

We move all terms to the left:

n^2+4n-7-(28)=0

We add all the numbers together, and all the variables

n^2+4n-35=0

a = 1; b = 4; c = -35;
Δ = b2-4ac
Δ = 42-4·1·(-35)
Δ = 156
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:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{39}}{2*1}=\frac{-4-2\sqrt{39}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{39}}{2*1}=\frac{-4+2\sqrt{39}}{2}
$