4y^2+10y+3=0

Solution for 4y^2+10y+3=0 equation:

4y^2+10y+3=0

a = 4; b = 10; c = +3;
Δ = b2-4ac
Δ = 102-4·4·3
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{13}}{2*4}=\frac{-10-2\sqrt{13}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{13}}{2*4}=\frac{-10+2\sqrt{13}}{8}
$