4w^2+12w+9=160

Solution for 4w^2+12w+9=160 equation:

4w^2+12w+9=160

We move all terms to the left:

4w^2+12w+9-(160)=0

We add all the numbers together, and all the variables

4w^2+12w-151=0

a = 4; b = 12; c = -151;
Δ = b2-4ac
Δ = 122-4·4·(-151)
Δ = 2560
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-16\sqrt{10}}{2*4}=\frac{-12-16\sqrt{10}}{8}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+16\sqrt{10}}{2*4}=\frac{-12+16\sqrt{10}}{8}
$