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2x^2+80=26x
We move all terms to the left:
2x^2+80-(26x)=0
a = 2; b = -26; c = +80;
Δ = b2-4ac
Δ = -262-4·2·80
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-6}{2*2}=\frac{20}{4} =5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+6}{2*2}=\frac{32}{4} =8 $
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