600x^2+20x-2=1

Solution for 600x^2+20x-2=1 equation:

600x^2+20x-2=1

We move all terms to the left:

600x^2+20x-2-(1)=0

We add all the numbers together, and all the variables

600x^2+20x-3=0

a = 600; b = 20; c = -3;
Δ = b2-4ac
Δ = 202-4·600·(-3)
Δ = 7600
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:

$\sqrt{\Delta}=\sqrt{7600}=\sqrt{400*19}=\sqrt{400}*\sqrt{19}=20\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{19}}{2*600}=\frac{-20-20\sqrt{19}}{1200}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{19}}{2*600}=\frac{-20+20\sqrt{19}}{1200}
$