2w2-14w+21=(w-3)2

Solution for 2w2-14w+21=(w-3)2 equation:

2w^2-14w+21=(w-3)2

We move all terms to the left:

2w^2-14w+21-((w-3)2)=0


We calculate terms in parentheses: -((w-3)2), so:

(w-3)2

We multiply parentheses

2w-6

Back to the equation:

-(2w-6)


We get rid of parentheses

2w^2-14w-2w+6+21=0

We add all the numbers together, and all the variables

2w^2-16w+27=0

a = 2; b = -16; c = +27;
Δ = b2-4ac
Δ = -162-4·2·27
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{10}}{2*2}=\frac{16-2\sqrt{10}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{10}}{2*2}=\frac{16+2\sqrt{10}}{4}
$