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49x^2+6x-2=0
a = 49; b = 6; c = -2;
Δ = b2-4ac
Δ = 62-4·49·(-2)
Δ = 428
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{428}=\sqrt{4*107}=\sqrt{4}*\sqrt{107}=2\sqrt{107}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{107}}{2*49}=\frac{-6-2\sqrt{107}}{98} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{107}}{2*49}=\frac{-6+2\sqrt{107}}{98} $
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