4y^2+6y=28

Solution for 4y^2+6y=28 equation:

4y^2+6y=28

We move all terms to the left:

4y^2+6y-(28)=0

a = 4; b = 6; c = -28;
Δ = b2-4ac
Δ = 62-4·4·(-28)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-22}{2*4}=\frac{-28}{8}
=-3+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+22}{2*4}=\frac{16}{8}
=2
$