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23x-5x^2=3x
We move all terms to the left:
23x-5x^2-(3x)=0
We add all the numbers together, and all the variables
-5x^2+20x=0
a = -5; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·(-5)·0
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*-5}=\frac{-40}{-10} =+4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*-5}=\frac{0}{-10} =0 $
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