10a+30a+40=a2+6a+4

Solution for 10a+30a+40=a2+6a+4 equation:

10a+30a+40=a2+6a+4

We move all terms to the left:

10a+30a+40-(a2+6a+4)=0

We add all the numbers together, and all the variables

-(+a^2+6a+4)+10a+30a+40=0

We add all the numbers together, and all the variables

-(+a^2+6a+4)+40a+40=0

We get rid of parentheses

-a^2-6a+40a-4+40=0

We add all the numbers together, and all the variables

-1a^2+34a+36=0

a = -1; b = 34; c = +36;
Δ = b2-4ac
Δ = 342-4·(-1)·36
Δ = 1300
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1300}=\sqrt{100*13}=\sqrt{100}*\sqrt{13}=10\sqrt{13}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-10\sqrt{13}}{2*-1}=\frac{-34-10\sqrt{13}}{-2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+10\sqrt{13}}{2*-1}=\frac{-34+10\sqrt{13}}{-2}
$