22w-4=-14+w2

Solution for 22w-4=-14+w2 equation:

22w-4=-14+w2

We move all terms to the left:

22w-4-(-14+w2)=0

We add all the numbers together, and all the variables

-(+w^2-14)+22w-4=0

We get rid of parentheses

-w^2+22w+14-4=0

We add all the numbers together, and all the variables

-1w^2+22w+10=0

a = -1; b = 22; c = +10;
Δ = b2-4ac
Δ = 222-4·(-1)·10
Δ = 524
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:

$\sqrt{\Delta}=\sqrt{524}=\sqrt{4*131}=\sqrt{4}*\sqrt{131}=2\sqrt{131}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{131}}{2*-1}=\frac{-22-2\sqrt{131}}{-2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{131}}{2*-1}=\frac{-22+2\sqrt{131}}{-2}
$