2x^2-2,25+3x+x^2=x^2+x+0,25

Solution for 2x^2-2,25+3x+x^2=x^2+x+0,25 equation:

2x^2-2.25+3x+x^2=x^2+x+0.25

We move all terms to the left:

2x^2-2.25+3x+x^2-(x^2+x+0.25)=0

We add all the numbers together, and all the variables

3x^2+3x-(x^2+x+0.25)-2.25=0

We get rid of parentheses

3x^2-x^2+3x-x-0.25-2.25=0

We add all the numbers together, and all the variables

2x^2+2x-2.5=0

a = 2; b = 2; c = -2.5;
Δ = b2-4ac
Δ = 22-4·2·(-2.5)
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{6}}{2*2}=\frac{-2-2\sqrt{6}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{6}}{2*2}=\frac{-2+2\sqrt{6}}{4}
$