10x^2+20x-100=38

Solution for 10x^2+20x-100=38 equation:

10x^2+20x-100=38

We move all terms to the left:

10x^2+20x-100-(38)=0

We add all the numbers together, and all the variables

10x^2+20x-138=0

a = 10; b = 20; c = -138;
Δ = b2-4ac
Δ = 202-4·10·(-138)
Δ = 5920
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{5920}=\sqrt{16*370}=\sqrt{16}*\sqrt{370}=4\sqrt{370}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{370}}{2*10}=\frac{-20-4\sqrt{370}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{370}}{2*10}=\frac{-20+4\sqrt{370}}{20}
$