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4u^2+16u-48=0
a = 4; b = 16; c = -48;
Δ = b2-4ac
Δ = 162-4·4·(-48)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1024}=32$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-32}{2*4}=\frac{-48}{8} =-6 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+32}{2*4}=\frac{16}{8} =2 $
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