3a(2a+4)+5a(7a-5)=

Solution for 3a(2a+4)+5a(7a-5)= equation:

3a(2a+4)+5a(7a-5)=

We move all terms to the left:

3a(2a+4)+5a(7a-5)-()=0

We add all the numbers together, and all the variables

3a(2a+4)+5a(7a-5)=0

We multiply parentheses

6a^2+35a^2+12a-25a=0

We add all the numbers together, and all the variables

41a^2-13a=0

a = 41; b = -13; c = 0;
Δ = b2-4ac
Δ = -132-4·41·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-13}{2*41}=\frac{0}{82}
=0
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+13}{2*41}=\frac{26}{82}
=13/41
$