4x^2=98

Solution for 4x^2=98 equation:

4x^2=98

We move all terms to the left:

4x^2-(98)=0

a = 4; b = 0; c = -98;
Δ = b2-4ac
Δ = 02-4·4·(-98)
Δ = 1568
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:

$\sqrt{\Delta}=\sqrt{1568}=\sqrt{784*2}=\sqrt{784}*\sqrt{2}=28\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-28\sqrt{2}}{2*4}=\frac{0-28\sqrt{2}}{8}
=-\frac{28\sqrt{2}}{8}
=-\frac{7\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+28\sqrt{2}}{2*4}=\frac{0+28\sqrt{2}}{8}
=\frac{28\sqrt{2}}{8}
=\frac{7\sqrt{2}}{2}
$