(2x^2+4x-1)/(5x-7)=0

Solution for (2x^2+4x-1)/(5x-7)=0 equation:

(2x^2+4x-1)/(5x-7)=0


Domain of the equation: (5x-7)!=0

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

5x!=7

x!=7/5

x!=1+2/5

x∈R


We multiply all the terms by the denominator


(2x^2+4x-1)=0

We get rid of parentheses

2x^2+4x-1=0

a = 2; b = 4; c = -1;
Δ = b2-4ac
Δ = 42-4·2·(-1)
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{6}}{2*2}=\frac{-4-2\sqrt{6}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{6}}{2*2}=\frac{-4+2\sqrt{6}}{4}
$