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49x^2-28x=4
We move all terms to the left:
49x^2-28x-(4)=0
a = 49; b = -28; c = -4;
Δ = b2-4ac
Δ = -282-4·49·(-4)
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{-(-28)-28\sqrt{2}}{2*49}=\frac{28-28\sqrt{2}}{98} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+28\sqrt{2}}{2*49}=\frac{28+28\sqrt{2}}{98} $
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