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6x^2-49x+100=0
a = 6; b = -49; c = +100;
Δ = b2-4ac
Δ = -492-4·6·100
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-49)-1}{2*6}=\frac{48}{12} =4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+1}{2*6}=\frac{50}{12} =4+1/6 $
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