a2-4a=3(a+20)

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Solution for a2-4a=3(a+20) equation:



a2-4a=3(a+20)
We move all terms to the left:
a2-4a-(3(a+20))=0
We add all the numbers together, and all the variables
a^2-4a-(3(a+20))=0
We calculate terms in parentheses: -(3(a+20)), so:
3(a+20)
We multiply parentheses
3a+60
Back to the equation:
-(3a+60)
We get rid of parentheses
a^2-4a-3a-60=0
We add all the numbers together, and all the variables
a^2-7a-60=0
a = 1; b = -7; c = -60;
Δ = b2-4ac
Δ = -72-4·1·(-60)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-17}{2*1}=\frac{-10}{2} =-5 $
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+17}{2*1}=\frac{24}{2} =12 $

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