4y^2+12y=7

Solution for 4y^2+12y=7 equation:

4y^2+12y=7

We move all terms to the left:

4y^2+12y-(7)=0

a = 4; b = 12; c = -7;
Δ = b2-4ac
Δ = 122-4·4·(-7)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-16}{2*4}=\frac{-28}{8}
=-3+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+16}{2*4}=\frac{4}{8}
=1/2
$