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2x^2-16x+64=40
We move all terms to the left:
2x^2-16x+64-(40)=0
We add all the numbers together, and all the variables
2x^2-16x+24=0
a = 2; b = -16; c = +24;
Δ = b2-4ac
Δ = -162-4·2·24
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8}{2*2}=\frac{8}{4} =2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8}{2*2}=\frac{24}{4} =6 $
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