(4x+4)(2x-1)-x^2+4=0

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Solution for (4x+4)(2x-1)-x^2+4=0 equation:



(4x+4)(2x-1)-x^2+4=0
We add all the numbers together, and all the variables
-1x^2+(4x+4)(2x-1)+4=0
We multiply parentheses ..
-1x^2+(+8x^2-4x+8x-4)+4=0
We get rid of parentheses
-1x^2+8x^2-4x+8x-4+4=0
We add all the numbers together, and all the variables
7x^2+4x=0
a = 7; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·7·0
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*7}=\frac{-8}{14} =-4/7 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*7}=\frac{0}{14} =0 $

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