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32+16x^2-48x=0
a = 16; b = -48; c = +32;
Δ = b2-4ac
Δ = -482-4·16·32
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-16}{2*16}=\frac{32}{32} =1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+16}{2*16}=\frac{64}{32} =2 $
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