4y^2+16y=7

Solution for 4y^2+16y=7 equation:

4y^2+16y=7

We move all terms to the left:

4y^2+16y-(7)=0

a = 4; b = 16; c = -7;
Δ = b2-4ac
Δ = 162-4·4·(-7)
Δ = 368
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:

$\sqrt{\Delta}=\sqrt{368}=\sqrt{16*23}=\sqrt{16}*\sqrt{23}=4\sqrt{23}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{23}}{2*4}=\frac{-16-4\sqrt{23}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{23}}{2*4}=\frac{-16+4\sqrt{23}}{8}
$