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5x^2-96+4x=0
a = 5; b = 4; c = -96;
Δ = b2-4ac
Δ = 42-4·5·(-96)
Δ = 1936
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{1936}=44$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-44}{2*5}=\frac{-48}{10} =-4+4/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+44}{2*5}=\frac{40}{10} =4 $
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