(2x^2+2x-4)/(5x+15)=0

Solution for (2x^2+2x-4)/(5x+15)=0 equation:

(2x^2+2x-4)/(5x+15)=0


Domain of the equation: (5x+15)!=0

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

5x!=-15

x!=-15/5

x!=-3

x∈R


We multiply all the terms by the denominator


(2x^2+2x-4)=0

We get rid of parentheses

2x^2+2x-4=0

a = 2; b = 2; c = -4;
Δ = b2-4ac
Δ = 22-4·2·(-4)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6}{2*2}=\frac{-8}{4}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6}{2*2}=\frac{4}{4}
=1
$