(8x-3)(2x+2)=0

Solution for (8x-3)(2x+2)=0 equation:

(8x-3)(2x+2)=0

We multiply parentheses ..

(+16x^2+16x-6x-6)=0

We get rid of parentheses

16x^2+16x-6x-6=0

We add all the numbers together, and all the variables

16x^2+10x-6=0

a = 16; b = 10; c = -6;
Δ = b2-4ac
Δ = 102-4·16·(-6)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-22}{2*16}=\frac{-32}{32}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+22}{2*16}=\frac{12}{32}
=3/8
$