(2w-5)(w+2)=-8

Solution for (2w-5)(w+2)=-8 equation:

(2w-5)(w+2)=-8

We move all terms to the left:

(2w-5)(w+2)-(-8)=0

We add all the numbers together, and all the variables

(2w-5)(w+2)+8=0

We multiply parentheses ..

(+2w^2+4w-5w-10)+8=0

We get rid of parentheses

2w^2+4w-5w-10+8=0

We add all the numbers together, and all the variables

2w^2-1w-2=0

a = 2; b = -1; c = -2;
Δ = b2-4ac
Δ = -12-4·2·(-2)
Δ = 17
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{-(-1)-\sqrt{17}}{2*2}=\frac{1-\sqrt{17}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{17}}{2*2}=\frac{1+\sqrt{17}}{4}
$