2a+4a-7a=6a+(4a)2

Solution for 2a+4a-7a=6a+(4a)2 equation:

2a+4a-7a=6a+(4a)2

We move all terms to the left:

2a+4a-7a-(6a+(4a)2)=0

We add all the numbers together, and all the variables

-(+6a+4a^2)+2a+4a-7a=0

We add all the numbers together, and all the variables

-(+6a+4a^2)-1a=0

We get rid of parentheses

-4a^2-6a-1a=0

We add all the numbers together, and all the variables

-4a^2-7a=0

a = -4; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·(-4)·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*-4}=\frac{0}{-8}
=0
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*-4}=\frac{14}{-8}
=-1+3/4
$