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10x^2+19x-10=0
a = 10; b = 19; c = -10;
Δ = b2-4ac
Δ = 192-4·10·(-10)
Δ = 761
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{-(19)-\sqrt{761}}{2*10}=\frac{-19-\sqrt{761}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{761}}{2*10}=\frac{-19+\sqrt{761}}{20} $
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