a2-4a=3(a+20)

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
$