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49x^2-28x-7=0
a = 49; b = -28; c = -7;
Δ = b2-4ac
Δ = -282-4·49·(-7)
Δ = 2156
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2156}=\sqrt{196*11}=\sqrt{196}*\sqrt{11}=14\sqrt{11}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-14\sqrt{11}}{2*49}=\frac{28-14\sqrt{11}}{98} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+14\sqrt{11}}{2*49}=\frac{28+14\sqrt{11}}{98} $
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