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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+2w^2-2w+5w-5)-2=0

We get rid of parentheses

2w^2-2w+5w-5-2=0

We add all the numbers together, and all the variables

2w^2+3w-7=0

a = 2; b = 3; c = -7;
Δ = b2-4ac
Δ = 32-4·2·(-7)
Δ = 65
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{65}}{2*2}=\frac{-3-\sqrt{65}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{65}}{2*2}=\frac{-3+\sqrt{65}}{4}
$