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a+5(2a-1)+3=11a^2
We move all terms to the left:
a+5(2a-1)+3-(11a^2)=0
determiningTheFunctionDomain -11a^2+a+5(2a-1)+3=0
We multiply parentheses
-11a^2+a+10a-5+3=0
We add all the numbers together, and all the variables
-11a^2+11a-2=0
a = -11; b = 11; c = -2;
Δ = b2-4ac
Δ = 112-4·(-11)·(-2)
Δ = 33
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{-(11)-\sqrt{33}}{2*-11}=\frac{-11-\sqrt{33}}{-22} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{33}}{2*-11}=\frac{-11+\sqrt{33}}{-22} $
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