2y2+y=10

Solution for 2y2+y=10 equation:

2y^2+y=10

We move all terms to the left:

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

a = 2; b = 1; c = -10;
Δ = b2-4ac
Δ = 12-4·2·(-10)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*2}=\frac{-10}{4}
=-2+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*2}=\frac{8}{4}
=2
$