5y-4=3/4y-7

Solution for 5y-4=3/4y-7 equation:

5y-4=3/4y-7

We move all terms to the left:

5y-4-(3/4y-7)=0


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

y∈R


We get rid of parentheses

5y-3/4y+7-4=0

We multiply all the terms by the denominator


5y*4y+7*4y-4*4y-3=0

Wy multiply elements

20y^2+28y-16y-3=0

We add all the numbers together, and all the variables

20y^2+12y-3=0

a = 20; b = 12; c = -3;
Δ = b2-4ac
Δ = 122-4·20·(-3)
Δ = 384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{6}}{2*20}=\frac{-12-8\sqrt{6}}{40}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{6}}{2*20}=\frac{-12+8\sqrt{6}}{40}
$