16x2+32=4(x+8)

Solution for 16x2+32=4(x+8) equation:

16x^2+32=4(x+8)

We move all terms to the left:

16x^2+32-(4(x+8))=0


We calculate terms in parentheses: -(4(x+8)), so:

4(x+8)

We multiply parentheses

4x+32

Back to the equation:

-(4x+32)


We get rid of parentheses

16x^2-4x-32+32=0

We add all the numbers together, and all the variables

16x^2-4x=0

a = 16; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·16·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*16}=\frac{0}{32}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*16}=\frac{8}{32}
=1/4
$