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12x+40+2x^2=360
We move all terms to the left:
12x+40+2x^2-(360)=0
We add all the numbers together, and all the variables
2x^2+12x-320=0
a = 2; b = 12; c = -320;
Δ = b2-4ac
Δ = 122-4·2·(-320)
Δ = 2704
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{2704}=52$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-52}{2*2}=\frac{-64}{4} =-16 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+52}{2*2}=\frac{40}{4} =10 $
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