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a^2+3=10a
We move all terms to the left:
a^2+3-(10a)=0
a = 1; b = -10; c = +3;
Δ = b2-4ac
Δ = -102-4·1·3
Δ = 88
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{22}}{2*1}=\frac{10-2\sqrt{22}}{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{22}}{2*1}=\frac{10+2\sqrt{22}}{2} $
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