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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

12a^2-15a+4=0

a = 12; b = -15; c = +4;
Δ = b2-4ac
Δ = -152-4·12·4
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{33}}{2*12}=\frac{15-\sqrt{33}}{24}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{33}}{2*12}=\frac{15+\sqrt{33}}{24}
$