98x^2+28x+1=0

Solution for 98x^2+28x+1=0 equation:

98x^2+28x+1=0

a = 98; b = 28; c = +1;
Δ = b2-4ac
Δ = 282-4·98·1
Δ = 392
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:

$\sqrt{\Delta}=\sqrt{392}=\sqrt{196*2}=\sqrt{196}*\sqrt{2}=14\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-14\sqrt{2}}{2*98}=\frac{-28-14\sqrt{2}}{196}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+14\sqrt{2}}{2*98}=\frac{-28+14\sqrt{2}}{196}
$