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5+19x-4x^2=0
a = -4; b = 19; c = +5;
Δ = b2-4ac
Δ = 192-4·(-4)·5
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-21}{2*-4}=\frac{-40}{-8} =+5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+21}{2*-4}=\frac{2}{-8} =-1/4 $
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