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2x^2+16x-4x-32=1008
We move all terms to the left:
2x^2+16x-4x-32-(1008)=0
We add all the numbers together, and all the variables
2x^2+12x-1040=0
a = 2; b = 12; c = -1040;
Δ = b2-4ac
Δ = 122-4·2·(-1040)
Δ = 8464
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{8464}=92$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-92}{2*2}=\frac{-104}{4} =-26 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+92}{2*2}=\frac{80}{4} =20 $
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