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20-5x^2=x
We move all terms to the left:
20-5x^2-(x)=0
We add all the numbers together, and all the variables
-5x^2-1x+20=0
a = -5; b = -1; c = +20;
Δ = b2-4ac
Δ = -12-4·(-5)·20
Δ = 401
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{401}}{2*-5}=\frac{1-\sqrt{401}}{-10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{401}}{2*-5}=\frac{1+\sqrt{401}}{-10} $
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