-245+100x-5x^2=0

Solution for -245+100x-5x^2=0 equation:

-245+100x-5x^2=0

a = -5; b = 100; c = -245;
Δ = b2-4ac
Δ = 1002-4·(-5)·(-245)
Δ = 5100
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:

$\sqrt{\Delta}=\sqrt{5100}=\sqrt{100*51}=\sqrt{100}*\sqrt{51}=10\sqrt{51}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-10\sqrt{51}}{2*-5}=\frac{-100-10\sqrt{51}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+10\sqrt{51}}{2*-5}=\frac{-100+10\sqrt{51}}{-10}
$