(14+3y)=(-5y-18)y

Solution for (14+3y)=(-5y-18)y equation:

(14+3y)=(-5y-18)y

We move all terms to the left:

(14+3y)-((-5y-18)y)=0

We add all the numbers together, and all the variables

(3y+14)-((-5y-18)y)=0

We get rid of parentheses

3y-((-5y-18)y)+14=0


We calculate terms in parentheses: -((-5y-18)y), so:

(-5y-18)y

We multiply parentheses

-5y^2-18y

Back to the equation:

-(-5y^2-18y)


We get rid of parentheses

5y^2+18y+3y+14=0

We add all the numbers together, and all the variables

5y^2+21y+14=0

a = 5; b = 21; c = +14;
Δ = b2-4ac
Δ = 212-4·5·14
Δ = 161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-\sqrt{161}}{2*5}=\frac{-21-\sqrt{161}}{10}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{161}}{2*5}=\frac{-21+\sqrt{161}}{10}
$