a^2+11a=4a+8

Solution for a^2+11a=4a+8 equation:

a^2+11a=4a+8

We move all terms to the left:

a^2+11a-(4a+8)=0

We get rid of parentheses

a^2+11a-4a-8=0

We add all the numbers together, and all the variables

a^2+7a-8=0

a = 1; b = 7; c = -8;
Δ = b2-4ac
Δ = 72-4·1·(-8)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-9}{2*1}=\frac{-16}{2}
=-8
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+9}{2*1}=\frac{2}{2}
=1
$