2n^2+20n-38=10

Solution for 2n^2+20n-38=10 equation:

2n^2+20n-38=10

We move all terms to the left:

2n^2+20n-38-(10)=0

We add all the numbers together, and all the variables

2n^2+20n-48=0

a = 2; b = 20; c = -48;
Δ = b2-4ac
Δ = 202-4·2·(-48)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-28}{2*2}=\frac{-48}{4}
=-12
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+28}{2*2}=\frac{8}{4}
=2
$