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19x^2+9x-3=0
a = 19; b = 9; c = -3;
Δ = b2-4ac
Δ = 92-4·19·(-3)
Δ = 309
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{-(9)-\sqrt{309}}{2*19}=\frac{-9-\sqrt{309}}{38} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{309}}{2*19}=\frac{-9+\sqrt{309}}{38} $
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