2*x^2=(2.5)^2-4*2.5*x+4*x^2

Solution for 2*x^2=(2.5)^2-4*2.5*x+4*x^2 equation:

2x^2=(2.5)^2-4*2.5x+4x^2

We move all terms to the left:

2x^2-((2.5)^2-4*2.5x+4x^2)=0


We calculate terms in parentheses: -((2.5)^2-4*2.5x+4x^2), so:

(2.5)^2-4*2.5x+4x^2

determiningTheFunctionDomain
4x^2-4*2.5x+(2.5)^2

We add all the numbers together, and all the variables

4x^2-4*2.5x+6.25

Wy multiply elements

4x^2-8x+6.25

Back to the equation:

-(4x^2-8x+6.25)


We get rid of parentheses

2x^2-4x^2+8x-6.25=0

We add all the numbers together, and all the variables

-2x^2+8x-6.25=0

a = -2; b = 8; c = -6.25;
Δ = b2-4ac
Δ = 82-4·(-2)·(-6.25)
Δ = 14
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-\sqrt{14}}{2*-2}=\frac{-8-\sqrt{14}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+\sqrt{14}}{2*-2}=\frac{-8+\sqrt{14}}{-4}
$