4y^2+10y=12

Solution for 4y^2+10y=12 equation:

4y^2+10y=12

We move all terms to the left:

4y^2+10y-(12)=0

a = 4; b = 10; c = -12;
Δ = b2-4ac
Δ = 102-4·4·(-12)
Δ = 292
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:

$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{73}}{2*4}=\frac{-10-2\sqrt{73}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{73}}{2*4}=\frac{-10+2\sqrt{73}}{8}
$