11y+2y^2=21

Solution for 11y+2y^2=21 equation:

11y+2y^2=21

We move all terms to the left:

11y+2y^2-(21)=0

a = 2; b = 11; c = -21;
Δ = b2-4ac
Δ = 112-4·2·(-21)
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{-(11)-17}{2*2}=\frac{-28}{4}
=-7
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+17}{2*2}=\frac{6}{4}
=1+1/2
$