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49x^2-56x+11=0
a = 49; b = -56; c = +11;
Δ = b2-4ac
Δ = -562-4·49·11
Δ = 980
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{980}=\sqrt{196*5}=\sqrt{196}*\sqrt{5}=14\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-56)-14\sqrt{5}}{2*49}=\frac{56-14\sqrt{5}}{98} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-56)+14\sqrt{5}}{2*49}=\frac{56+14\sqrt{5}}{98} $
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