(5/4)(4x+2)=3

Solution for (5/4)(4x+2)=3 equation:

(5/4)(4x+2)=3

We move all terms to the left:

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


Domain of the equation: 4)(4x+2)!=0

x∈R


We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+20x^2+5/4*2)-3=0

We multiply all the terms by the denominator


(+20x^2+5-3*4*2)=0

We get rid of parentheses

20x^2+5-3*4*2=0

We add all the numbers together, and all the variables

20x^2-19=0

a = 20; b = 0; c = -19;
Δ = b2-4ac
Δ = 02-4·20·(-19)
Δ = 1520
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:

$\sqrt{\Delta}=\sqrt{1520}=\sqrt{16*95}=\sqrt{16}*\sqrt{95}=4\sqrt{95}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{95}}{2*20}=\frac{0-4\sqrt{95}}{40}
=-\frac{4\sqrt{95}}{40}
=-\frac{\sqrt{95}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{95}}{2*20}=\frac{0+4\sqrt{95}}{40}
=\frac{4\sqrt{95}}{40}
=\frac{\sqrt{95}}{10}
$