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300=-0.5x^2+40x-300
We move all terms to the left:
300-(-0.5x^2+40x-300)=0
We get rid of parentheses
0.5x^2-40x+300+300=0
We add all the numbers together, and all the variables
0.5x^2-40x+600=0
a = 0.5; b = -40; c = +600;
Δ = b2-4ac
Δ = -402-4·0.5·600
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-20}{2*0.5}=\frac{20}{1} =20 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+20}{2*0.5}=\frac{60}{1} =60 $
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