3+(2y-1)(y+5)=(y+2)(y+7)

Solution for 3+(2y-1)(y+5)=(y+2)(y+7) equation:

3+(2y-1)(y+5)=(y+2)(y+7)

We move all terms to the left:

3+(2y-1)(y+5)-((y+2)(y+7))=0

We multiply parentheses ..

(+2y^2+10y-1y-5)-((y+2)(y+7))+3=0


We calculate terms in parentheses: -((y+2)(y+7)), so:

(y+2)(y+7)

We multiply parentheses ..

(+y^2+7y+2y+14)

We get rid of parentheses

y^2+7y+2y+14

We add all the numbers together, and all the variables

y^2+9y+14

Back to the equation:

-(y^2+9y+14)


We get rid of parentheses

2y^2-y^2+10y-1y-9y-5-14+3=0

We add all the numbers together, and all the variables

y^2-16=0

a = 1; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·1·(-16)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*1}=\frac{-8}{2}
=-4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*1}=\frac{8}{2}
=4
$