0=4(x+3)(x-5)

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

0=4(x+3)(x-5)

We move all terms to the left:

0-(4(x+3)(x-5))=0

We add all the numbers together, and all the variables

-(4(x+3)(x-5))=0

We multiply parentheses ..

-(4(+x^2-5x+3x-15))=0


We calculate terms in parentheses: -(4(+x^2-5x+3x-15)), so:

4(+x^2-5x+3x-15)

We multiply parentheses

4x^2-20x+12x-60

We add all the numbers together, and all the variables

4x^2-8x-60

Back to the equation:

-(4x^2-8x-60)


We get rid of parentheses

-4x^2+8x+60=0

a = -4; b = 8; c = +60;
Δ = b2-4ac
Δ = 82-4·(-4)·60
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-32}{2*-4}=\frac{-40}{-8}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+32}{2*-4}=\frac{24}{-8}
=-3
$