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5x^2+19x-4=0
a = 5; b = 19; c = -4;
Δ = b2-4ac
Δ = 192-4·5·(-4)
Δ = 441
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{441}=21$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-21}{2*5}=\frac{-40}{10} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+21}{2*5}=\frac{2}{10} =1/5 $
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