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4x^2+28x-120=0
a = 4; b = 28; c = -120;
Δ = b2-4ac
Δ = 282-4·4·(-120)
Δ = 2704
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{2704}=52$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-52}{2*4}=\frac{-80}{8} =-10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+52}{2*4}=\frac{24}{8} =3 $
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