((2x-3)(x+5))/(x-7)=0

Solution for ((2x-3)(x+5))/(x-7)=0 equation:

((2x-3)(x+5))/(x-7)=0


Domain of the equation: (x-7)!=0

We move all terms containing x to the left, all other terms to the right

x!=7

x∈R


We multiply parentheses ..

((+2x^2+10x-3x-15))/(x-7)=0

We multiply all the terms by the denominator


((+2x^2+10x-3x-15))=0


We calculate terms in parentheses: +((+2x^2+10x-3x-15)), so:

(+2x^2+10x-3x-15)

We get rid of parentheses

2x^2+10x-3x-15

We add all the numbers together, and all the variables

2x^2+7x-15

Back to the equation:

+(2x^2+7x-15)


We get rid of parentheses

2x^2+7x-15=0

a = 2; b = 7; c = -15;
Δ = b2-4ac
Δ = 72-4·2·(-15)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-13}{2*2}=\frac{-20}{4}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+13}{2*2}=\frac{6}{4}
=1+1/2
$