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4x^2-32x+48=0
a = 4; b = -32; c = +48;
Δ = b2-4ac
Δ = -322-4·4·48
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16}{2*4}=\frac{16}{8} =2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16}{2*4}=\frac{48}{8} =6 $
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