x^2+(4-x/2)^2=40

Solution for x^2+(4-x/2)^2=40 equation:

x^2+(4-x/2)^2=40

We move all terms to the left:

x^2+(4-x/2)^2-(40)=0

We add all the numbers together, and all the variables

x^2+(-x/2+4)^2-40=0

We multiply all the terms by the denominator


x^2*2-x+4)^2+(-40*2+4)^2=0

We add all the numbers together, and all the variables

x^2*2-x+4)^2+(-76)^2=0

We add all the numbers together, and all the variables

x^2*2-1x=0

Wy multiply elements

2x^2-1x=0

a = 2; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·2·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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