(7y-4)(2y+3)-13y=0

Solution for (7y-4)(2y+3)-13y=0 equation:

(7y-4)(2y+3)-13y=0

We add all the numbers together, and all the variables

-13y+(7y-4)(2y+3)=0

We multiply parentheses ..

(+14y^2+21y-8y-12)-13y=0

We get rid of parentheses

14y^2+21y-8y-13y-12=0

We add all the numbers together, and all the variables

14y^2-12=0

a = 14; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·14·(-12)
Δ = 672
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:

$\sqrt{\Delta}=\sqrt{672}=\sqrt{16*42}=\sqrt{16}*\sqrt{42}=4\sqrt{42}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{42}}{2*14}=\frac{0-4\sqrt{42}}{28}
=-\frac{4\sqrt{42}}{28}
=-\frac{\sqrt{42}}{7}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{42}}{2*14}=\frac{0+4\sqrt{42}}{28}
=\frac{4\sqrt{42}}{28}
=\frac{\sqrt{42}}{7}
$