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x^2-20x=20
We move all terms to the left:
x^2-20x-(20)=0
a = 1; b = -20; c = -20;
Δ = b2-4ac
Δ = -202-4·1·(-20)
Δ = 480
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{30}}{2*1}=\frac{20-4\sqrt{30}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{30}}{2*1}=\frac{20+4\sqrt{30}}{2} $
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