(w+2)(2w-1)=(w-2)(w-5)+15

Solution for (w+2)(2w-1)=(w-2)(w-5)+15 equation:

(w+2)(2w-1)=(w-2)(w-5)+15

We move all terms to the left:

(w+2)(2w-1)-((w-2)(w-5)+15)=0

We multiply parentheses ..

(+2w^2-1w+4w-2)-((w-2)(w-5)+15)=0


We calculate terms in parentheses: -((w-2)(w-5)+15), so:

(w-2)(w-5)+15

We multiply parentheses ..

(+w^2-5w-2w+10)+15

We get rid of parentheses

w^2-5w-2w+10+15

We add all the numbers together, and all the variables

w^2-7w+25

Back to the equation:

-(w^2-7w+25)


We get rid of parentheses

2w^2-w^2-1w+4w+7w-2-25=0

We add all the numbers together, and all the variables

w^2+10w-27=0

a = 1; b = 10; c = -27;
Δ = b2-4ac
Δ = 102-4·1·(-27)
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{13}}{2*1}=\frac{-10-4\sqrt{13}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{13}}{2*1}=\frac{-10+4\sqrt{13}}{2}
$