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320-96x+4x^2=0
a = 4; b = -96; c = +320;
Δ = b2-4ac
Δ = -962-4·4·320
Δ = 4096
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{4096}=64$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-64}{2*4}=\frac{32}{8} =4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+64}{2*4}=\frac{160}{8} =20 $
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