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-0.5x^2+40x+250=50
We move all terms to the left:
-0.5x^2+40x+250-(50)=0
We add all the numbers together, and all the variables
-0.5x^2+40x+200=0
a = -0.5; b = 40; c = +200;
Δ = b2-4ac
Δ = 402-4·(-0.5)·200
Δ = 2000
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2000}=\sqrt{400*5}=\sqrt{400}*\sqrt{5}=20\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-20\sqrt{5}}{2*-0.5}=\frac{-40-20\sqrt{5}}{-1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20\sqrt{5}}{2*-0.5}=\frac{-40+20\sqrt{5}}{-1} $
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