(x^2+5x)/(4x-7)=0

Solution for (x^2+5x)/(4x-7)=0 equation:

(x^2+5x)/(4x-7)=0


Domain of the equation: (4x-7)!=0

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

4x!=7

x!=7/4

x!=1+3/4

x∈R


We multiply all the terms by the denominator


(x^2+5x)=0

We get rid of parentheses

x^2+5x=0

a = 1; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·1·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*1}=\frac{-10}{2}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*1}=\frac{0}{2}
=0
$