2=140/2+12-4x2

Solution for 2=140/2+12-4x2 equation:

2=140/2+12-4x^2

We move all terms to the left:

2-(140/2+12-4x^2)=0


Domain of the equation: 2+12-4x^2)!=0

We move all terms containing x to the left, all other terms to the right

-4x^2)!=-14

x!=-14/1

x!=-14

x∈R


We get rid of parentheses

4x^2-12+2-140/2=0

We multiply all the terms by the denominator


4x^2*2-140-12*2+2*2=0

We add all the numbers together, and all the variables

4x^2*2-160=0

Wy multiply elements

8x^2-160=0

a = 8; b = 0; c = -160;
Δ = b2-4ac
Δ = 02-4·8·(-160)
Δ = 5120
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{5120}=\sqrt{1024*5}=\sqrt{1024}*\sqrt{5}=32\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{5}}{2*8}=\frac{0-32\sqrt{5}}{16}
=-\frac{32\sqrt{5}}{16}
=-2\sqrt{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{5}}{2*8}=\frac{0+32\sqrt{5}}{16}
=\frac{32\sqrt{5}}{16}
=2\sqrt{5}
$