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16x^2-112x+192=0
a = 16; b = -112; c = +192;
Δ = b2-4ac
Δ = -1122-4·16·192
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-112)-16}{2*16}=\frac{96}{32} =3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-112)+16}{2*16}=\frac{128}{32} =4 $
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