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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

3(2a+5)+2(-1a+3)a-4=0

We multiply parentheses

-2a^2+6a+6a+15-4=0

We add all the numbers together, and all the variables

-2a^2+12a+11=0

a = -2; b = 12; c = +11;
Δ = b2-4ac
Δ = 122-4·(-2)·11
Δ = 232
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:

$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{58}}{2*-2}=\frac{-12-2\sqrt{58}}{-4}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{58}}{2*-2}=\frac{-12+2\sqrt{58}}{-4}
$