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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+2y^2+6y+5y+15)-(-5(2y+5))=0


We calculate terms in parentheses: -(-5(2y+5)), so:

-5(2y+5)

We multiply parentheses

-10y-25

Back to the equation:

-(-10y-25)


We get rid of parentheses

2y^2+6y+5y+10y+15+25=0

We add all the numbers together, and all the variables

2y^2+21y+40=0

a = 2; b = 21; c = +40;
Δ = b2-4ac
Δ = 212-4·2·40
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-11}{2*2}=\frac{-32}{4}
=-8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+11}{2*2}=\frac{-10}{4}
=-2+1/2
$