4x^2-2x-98=0

Solution for 4x^2-2x-98=0 equation:

4x^2-2x-98=0

a = 4; b = -2; c = -98;
Δ = b2-4ac
Δ = -22-4·4·(-98)
Δ = 1572
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:

$\sqrt{\Delta}=\sqrt{1572}=\sqrt{4*393}=\sqrt{4}*\sqrt{393}=2\sqrt{393}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{393}}{2*4}=\frac{2-2\sqrt{393}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{393}}{2*4}=\frac{2+2\sqrt{393}}{8}
$