2x^2-98x+5=0

Solution for 2x^2-98x+5=0 equation:

2x^2-98x+5=0

a = 2; b = -98; c = +5;
Δ = b2-4ac
Δ = -982-4·2·5
Δ = 9564
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:

$\sqrt{\Delta}=\sqrt{9564}=\sqrt{4*2391}=\sqrt{4}*\sqrt{2391}=2\sqrt{2391}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-98)-2\sqrt{2391}}{2*2}=\frac{98-2\sqrt{2391}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-98)+2\sqrt{2391}}{2*2}=\frac{98+2\sqrt{2391}}{4}
$