2y2+8y=10

Solution for 2y2+8y=10 equation:

2y^2+8y=10

We move all terms to the left:

2y^2+8y-(10)=0

a = 2; b = 8; c = -10;
Δ = b2-4ac
Δ = 82-4·2·(-10)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-12}{2*2}=\frac{-20}{4}
=-5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+12}{2*2}=\frac{4}{4}
=1
$