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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} $
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