3y2+7y+1=1

Solution for 3y2+7y+1=1 equation:

3y^2+7y+1=1

We move all terms to the left:

3y^2+7y+1-(1)=0

We add all the numbers together, and all the variables

3y^2+7y=0

a = 3; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·3·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*3}=\frac{-14}{6}
=-2+1/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*3}=\frac{0}{6}
=0
$