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

Solution for (4y-3)(2y+5y)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+28y^2-21y)=0

We get rid of parentheses

28y^2-21y=0

a = 28; b = -21; c = 0;
Δ = b2-4ac
Δ = -212-4·28·0
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-21}{2*28}=\frac{0}{56}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+21}{2*28}=\frac{42}{56}
=3/4
$