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a^2=3a+8
We move all terms to the left:
a^2-(3a+8)=0
We get rid of parentheses
a^2-3a-8=0
a = 1; b = -3; c = -8;
Δ = b2-4ac
Δ = -32-4·1·(-8)
Δ = 41
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{41}}{2*1}=\frac{3-\sqrt{41}}{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{41}}{2*1}=\frac{3+\sqrt{41}}{2} $
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