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49x^2-42x+8=0
a = 49; b = -42; c = +8;
Δ = b2-4ac
Δ = -422-4·49·8
Δ = 196
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-14}{2*49}=\frac{28}{98} =2/7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+14}{2*49}=\frac{56}{98} =4/7 $
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