(4x^2-3)/(5x-5)=0

Solution for (4x^2-3)/(5x-5)=0 equation:

(4x^2-3)/(5x-5)=0


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

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

5x!=5

x!=5/5

x!=1

x∈R


We multiply all the terms by the denominator


(4x^2-3)=0

We get rid of parentheses

4x^2-3=0

a = 4; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·4·(-3)
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*4}=\frac{0-4\sqrt{3}}{8}
=-\frac{4\sqrt{3}}{8}
=-\frac{\sqrt{3}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*4}=\frac{0+4\sqrt{3}}{8}
=\frac{4\sqrt{3}}{8}
=\frac{\sqrt{3}}{2}
$