81x^2+180x-200=100

Solution for 81x^2+180x-200=100 equation:

81x^2+180x-200=100

We move all terms to the left:

81x^2+180x-200-(100)=0

We add all the numbers together, and all the variables

81x^2+180x-300=0

a = 81; b = 180; c = -300;
Δ = b2-4ac
Δ = 1802-4·81·(-300)
Δ = 129600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{129600}=360$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(180)-360}{2*81}=\frac{-540}{162}
=-3+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(180)+360}{2*81}=\frac{180}{162}
=1+1/9
$