35=y(2y+3)

Solution for 35=y(2y+3) equation:

35=y(2y+3)

We move all terms to the left:

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


We calculate terms in parentheses: -(y(2y+3)), so:

y(2y+3)

We multiply parentheses

2y^2+3y

Back to the equation:

-(2y^2+3y)


We get rid of parentheses

-2y^2-3y+35=0

a = -2; b = -3; c = +35;
Δ = b2-4ac
Δ = -32-4·(-2)·35
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{-(-3)-17}{2*-2}=\frac{-14}{-4}
=3+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+17}{2*-2}=\frac{20}{-4}
=-5
$