X^2+48+10x+19=180

Solution for X^2+48+10x+19=180 equation:

X^2+48+10X+19=180

We move all terms to the left:

X^2+48+10X+19-(180)=0

We add all the numbers together, and all the variables

X^2+10X-113=0

a = 1; b = 10; c = -113;
Δ = b2-4ac
Δ = 102-4·1·(-113)
Δ = 552
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:

$\sqrt{\Delta}=\sqrt{552}=\sqrt{4*138}=\sqrt{4}*\sqrt{138}=2\sqrt{138}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{138}}{2*1}=\frac{-10-2\sqrt{138}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{138}}{2*1}=\frac{-10+2\sqrt{138}}{2}
$