4y^2+18y-14=0

Solution for 4y^2+18y-14=0 equation:

4y^2+18y-14=0

a = 4; b = 18; c = -14;
Δ = b2-4ac
Δ = 182-4·4·(-14)
Δ = 548
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{548}=\sqrt{4*137}=\sqrt{4}*\sqrt{137}=2\sqrt{137}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{137}}{2*4}=\frac{-18-2\sqrt{137}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{137}}{2*4}=\frac{-18+2\sqrt{137}}{8}
$