4y2+52y+162=0

Solution for 4y2+52y+162=0 equation:

4y^2+52y+162=0

a = 4; b = 52; c = +162;
Δ = b2-4ac
Δ = 522-4·4·162
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(52)-4\sqrt{7}}{2*4}=\frac{-52-4\sqrt{7}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(52)+4\sqrt{7}}{2*4}=\frac{-52+4\sqrt{7}}{8}
$