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300=40x-x^2
We move all terms to the left:
300-(40x-x^2)=0
We get rid of parentheses
x^2-40x+300=0
a = 1; b = -40; c = +300;
Δ = b2-4ac
Δ = -402-4·1·300
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-20}{2*1}=\frac{20}{2} =10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+20}{2*1}=\frac{60}{2} =30 $
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