(2x-5)(10x+3)=0

Solution for (2x-5)(10x+3)=0 equation:

(2x-5)(10x+3)=0

We multiply parentheses ..

(+20x^2+6x-50x-15)=0

We get rid of parentheses

20x^2+6x-50x-15=0

We add all the numbers together, and all the variables

20x^2-44x-15=0

a = 20; b = -44; c = -15;
Δ = b2-4ac
Δ = -442-4·20·(-15)
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-44)-56}{2*20}=\frac{-12}{40}
=-3/10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-44)+56}{2*20}=\frac{100}{40}
=2+1/2
$