(7y+4)(4y-5)=0

Solution for (7y+4)(4y-5)=0 equation:

(7y+4)(4y-5)=0

We multiply parentheses ..

(+28y^2-35y+16y-20)=0

We get rid of parentheses

28y^2-35y+16y-20=0

We add all the numbers together, and all the variables

28y^2-19y-20=0

a = 28; b = -19; c = -20;
Δ = b2-4ac
Δ = -192-4·28·(-20)
Δ = 2601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2601}=51$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-51}{2*28}=\frac{-32}{56}
=-4/7
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+51}{2*28}=\frac{70}{56}
=1+1/4
$