(2w+3)2+(w2)=78

Solution for (2w+3)2+(w2)=78 equation:

(2w+3)2+(w2)=78

We move all terms to the left:

(2w+3)2+(w2)-(78)=0

We add all the numbers together, and all the variables

w^2+(2w+3)2-78=0

We multiply parentheses

w^2+4w+6-78=0

We add all the numbers together, and all the variables

w^2+4w-72=0

a = 1; b = 4; c = -72;
Δ = b2-4ac
Δ = 42-4·1·(-72)
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{19}}{2*1}=\frac{-4-4\sqrt{19}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{19}}{2*1}=\frac{-4+4\sqrt{19}}{2}
$