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X2+20=10X
We move all terms to the left:
X2+20-(10X)=0
We add all the numbers together, and all the variables
X^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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