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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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16}{2*4}=\frac{-48}{8} =-6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16}{2*4}=\frac{-16}{8} =-2 $
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