3a2+4a=1

Solution for 3a2+4a=1 equation:

3a^2+4a=1

We move all terms to the left:

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

a = 3; b = 4; c = -1;
Δ = b2-4ac
Δ = 42-4·3·(-1)
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{7}}{2*3}=\frac{-4-2\sqrt{7}}{6}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{7}}{2*3}=\frac{-4+2\sqrt{7}}{6}
$