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7x^2+19x-4=0
a = 7; b = 19; c = -4;
Δ = b2-4ac
Δ = 192-4·7·(-4)
Δ = 473
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{473}}{2*7}=\frac{-19-\sqrt{473}}{14} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{473}}{2*7}=\frac{-19+\sqrt{473}}{14} $
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