9x2+-245x+500=0

Solution for 9x2+-245x+500=0 equation:

9x^2+-245x+500=0

We add all the numbers together, and all the variables

9x^2-245x=0

a = 9; b = -245; c = 0;
Δ = b2-4ac
Δ = -2452-4·9·0
Δ = 60025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{60025}=245$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-245)-245}{2*9}=\frac{0}{18}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-245)+245}{2*9}=\frac{490}{18}
=27+2/9
$