10x+x^2-4=99

Solution for 10x+x^2-4=99 equation:

10x+x^2-4=99

We move all terms to the left:

10x+x^2-4-(99)=0

We add all the numbers together, and all the variables

x^2+10x-103=0

a = 1; b = 10; c = -103;
Δ = b2-4ac
Δ = 102-4·1·(-103)
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-16\sqrt{2}}{2*1}=\frac{-10-16\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+16\sqrt{2}}{2*1}=\frac{-10+16\sqrt{2}}{2}
$