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4x^2+20x-96=0
a = 4; b = 20; c = -96;
Δ = b2-4ac
Δ = 202-4·4·(-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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-44}{2*4}=\frac{-64}{8} =-8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+44}{2*4}=\frac{24}{8} =3 $
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