(y2)+(3y+10)=90

Solution for (y2)+(3y+10)=90 equation:

(y2)+(3y+10)=90

We move all terms to the left:

(y2)+(3y+10)-(90)=0

We add all the numbers together, and all the variables

y^2+(3y+10)-90=0

We get rid of parentheses

y^2+3y+10-90=0

We add all the numbers together, and all the variables

y^2+3y-80=0

a = 1; b = 3; c = -80;
Δ = b2-4ac
Δ = 32-4·1·(-80)
Δ = 329
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{-(3)-\sqrt{329}}{2*1}=\frac{-3-\sqrt{329}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{329}}{2*1}=\frac{-3+\sqrt{329}}{2}
$