5y(2y-3)+(2y-3)=y+3/4

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

5y(2y-3)+(2y-3)=y+3/4

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

10y^2-15y+(2y-3)-(+y+3/4)=0

We get rid of parentheses

10y^2-15y+2y-y-3-3/4=0

We multiply all the terms by the denominator


10y^2*4-15y*4+2y*4-y*4-3-3*4=0

We add all the numbers together, and all the variables

10y^2*4-15y*4+2y*4-y*4-15=0

Wy multiply elements

40y^2-60y+8y-4y-15=0

We add all the numbers together, and all the variables

40y^2-56y-15=0

a = 40; b = -56; c = -15;
Δ = b2-4ac
Δ = -562-4·40·(-15)
Δ = 5536
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:

$\sqrt{\Delta}=\sqrt{5536}=\sqrt{16*346}=\sqrt{16}*\sqrt{346}=4\sqrt{346}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-56)-4\sqrt{346}}{2*40}=\frac{56-4\sqrt{346}}{80}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-56)+4\sqrt{346}}{2*40}=\frac{56+4\sqrt{346}}{80}
$