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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((+2x^2-7x+10x-35)), so:

(+2x^2-7x+10x-35)

We get rid of parentheses

2x^2-7x+10x-35

We add all the numbers together, and all the variables

2x^2+3x-35

Back to the equation:

-(2x^2+3x-35)


We get rid of parentheses

-2x^2-3x+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:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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