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(e-4)(e-10)=0
We multiply parentheses ..
(+e^2-10e-4e+40)=0
We get rid of parentheses
e^2-10e-4e+40=0
We add all the numbers together, and all the variables
e^2-14e+40=0
a = 1; b = -14; c = +40;
Δ = b2-4ac
Δ = -142-4·1·40
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{36}=6$$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-6}{2*1}=\frac{8}{2} =4 $$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+6}{2*1}=\frac{20}{2} =10 $
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