(2w-7)(w+5)=0

Solution for (2w-7)(w+5)=0 equation:

(2w-7)(w+5)=0

We multiply parentheses ..

(+2w^2+10w-7w-35)=0

We get rid of parentheses

2w^2+10w-7w-35=0

We add all the numbers together, and all the variables

2w^2+3w-35=0

a = 2; b = 3; c = -35;
Δ = b2-4ac
Δ = 32-4·2·(-35)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-17}{2*2}=\frac{-20}{4}
=-5
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+17}{2*2}=\frac{14}{4}
=3+1/2
$