3w+w^2+1=10-w

Solution for 3w+w^2+1=10-w equation:

3w+w^2+1=10-w

We move all terms to the left:

3w+w^2+1-(10-w)=0

We add all the numbers together, and all the variables

w^2+3w-(-1w+10)+1=0

We get rid of parentheses

w^2+3w+1w-10+1=0

We add all the numbers together, and all the variables

w^2+4w-9=0

a = 1; b = 4; c = -9;
Δ = b2-4ac
Δ = 42-4·1·(-9)
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{13}}{2*1}=\frac{-4-2\sqrt{13}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{13}}{2*1}=\frac{-4+2\sqrt{13}}{2}
$