482x+100x^2-5=0

Solution for 482x+100x^2-5=0 equation:

482x+100x^2-5=0

a = 100; b = 482; c = -5;
Δ = b2-4ac
Δ = 4822-4·100·(-5)
Δ = 234324
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:

$\sqrt{\Delta}=\sqrt{234324}=\sqrt{36*6509}=\sqrt{36}*\sqrt{6509}=6\sqrt{6509}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(482)-6\sqrt{6509}}{2*100}=\frac{-482-6\sqrt{6509}}{200}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(482)+6\sqrt{6509}}{2*100}=\frac{-482+6\sqrt{6509}}{200}
$