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2X^2-4X+20=180
We move all terms to the left:
2X^2-4X+20-(180)=0
We add all the numbers together, and all the variables
2X^2-4X-160=0
a = 2; b = -4; c = -160;
Δ = b2-4ac
Δ = -42-4·2·(-160)
Δ = 1296
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{1296}=36$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-36}{2*2}=\frac{-32}{4} =-8 $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+36}{2*2}=\frac{40}{4} =10 $
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