(2y-7)(y+3)=3+12

Solution for (2y-7)(y+3)=3+12 equation:

(2y-7)(y+3)=3+12

We move all terms to the left:

(2y-7)(y+3)-(3+12)=0

We add all the numbers together, and all the variables

(2y-7)(y+3)-15=0

We multiply parentheses ..

(+2y^2+6y-7y-21)-15=0

We get rid of parentheses

2y^2+6y-7y-21-15=0

We add all the numbers together, and all the variables

2y^2-1y-36=0

a = 2; b = -1; c = -36;
Δ = b2-4ac
Δ = -12-4·2·(-36)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-17}{2*2}=\frac{-16}{4}
=-4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+17}{2*2}=\frac{18}{4}
=4+1/2
$