4y2+18y=10

Solution for 4y2+18y=10 equation:

4y^2+18y=10

We move all terms to the left:

4y^2+18y-(10)=0

a = 4; b = 18; c = -10;
Δ = b2-4ac
Δ = 182-4·4·(-10)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-22}{2*4}=\frac{-40}{8}
=-5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+22}{2*4}=\frac{4}{8}
=1/2
$