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10x^2+49x+6=0
a = 10; b = 49; c = +6;
Δ = b2-4ac
Δ = 492-4·10·6
Δ = 2161
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-\sqrt{2161}}{2*10}=\frac{-49-\sqrt{2161}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+\sqrt{2161}}{2*10}=\frac{-49+\sqrt{2161}}{20} $
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