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

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

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

We multiply parentheses ..

(+16x^2-12x+20x-15)=0

We get rid of parentheses

16x^2-12x+20x-15=0

We add all the numbers together, and all the variables

16x^2+8x-15=0

a = 16; b = 8; c = -15;
Δ = b2-4ac
Δ = 82-4·16·(-15)
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*16}=\frac{-40}{32}
=-1+1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+32}{2*16}=\frac{24}{32}
=3/4
$