(4x+4)(2x-5)=0

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

(4x+4)(2x-5)=0

We multiply parentheses ..

(+8x^2-20x+8x-20)=0

We get rid of parentheses

8x^2-20x+8x-20=0

We add all the numbers together, and all the variables

8x^2-12x-20=0

a = 8; b = -12; c = -20;
Δ = b2-4ac
Δ = -122-4·8·(-20)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-28}{2*8}=\frac{-16}{16}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+28}{2*8}=\frac{40}{16}
=2+1/2
$