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10x-x^2=20
We move all terms to the left:
10x-x^2-(20)=0
We add all the numbers together, and all the variables
-1x^2+10x-20=0
a = -1; b = 10; c = -20;
Δ = b2-4ac
Δ = 102-4·(-1)·(-20)
Δ = 20
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{5}}{2*-1}=\frac{-10-2\sqrt{5}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{5}}{2*-1}=\frac{-10+2\sqrt{5}}{-2} $
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