Y=2x+3/4x+5

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

=2Y+3/4Y+5

We move all terms to the left:

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


Domain of the equation: 4Y+5)!=0

Y∈R


We get rid of parentheses

-2Y-3/4Y-5=0

We multiply all the terms by the denominator


-2Y*4Y-5*4Y-3=0

Wy multiply elements

-8Y^2-20Y-3=0

a = -8; b = -20; c = -3;
Δ = b2-4ac
Δ = -202-4·(-8)·(-3)
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{19}}{2*-8}=\frac{20-4\sqrt{19}}{-16}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{19}}{2*-8}=\frac{20+4\sqrt{19}}{-16}
$