4(2x^2+5x)=-8x+16

Solution for 4(2x^2+5x)=-8x+16 equation:

4(2x^2+5x)=-8x+16

We move all terms to the left:

4(2x^2+5x)-(-8x+16)=0

We multiply parentheses

8x^2+20x-(-8x+16)=0

We get rid of parentheses

8x^2+20x+8x-16=0

We add all the numbers together, and all the variables

8x^2+28x-16=0

a = 8; b = 28; c = -16;
Δ = b2-4ac
Δ = 282-4·8·(-16)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-36}{2*8}=\frac{-64}{16}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+36}{2*8}=\frac{8}{16}
=1/2
$