4y2+13y=12

Solution for 4y2+13y=12 equation:

4y^2+13y=12

We move all terms to the left:

4y^2+13y-(12)=0

a = 4; b = 13; c = -12;
Δ = b2-4ac
Δ = 132-4·4·(-12)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-19}{2*4}=\frac{-32}{8}
=-4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+19}{2*4}=\frac{6}{8}
=3/4
$