2y^2+23y+4=16

Solution for 2y^2+23y+4=16 equation:

2y^2+23y+4=16

We move all terms to the left:

2y^2+23y+4-(16)=0

We add all the numbers together, and all the variables

2y^2+23y-12=0

a = 2; b = 23; c = -12;
Δ = b2-4ac
Δ = 232-4·2·(-12)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-25}{2*2}=\frac{-48}{4}
=-12
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+25}{2*2}=\frac{2}{4}
=1/2
$