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(a-1)(a-3)=35
We move all terms to the left:
(a-1)(a-3)-(35)=0
We multiply parentheses ..
(+a^2-3a-1a+3)-35=0
We get rid of parentheses
a^2-3a-1a+3-35=0
We add all the numbers together, and all the variables
a^2-4a-32=0
a = 1; b = -4; c = -32;
Δ = b2-4ac
Δ = -42-4·1·(-32)
Δ = 144
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{144}=12$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-12}{2*1}=\frac{-8}{2} =-4 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12}{2*1}=\frac{16}{2} =8 $
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