(2x2+6x+10)+(=4x2+2x+3)

Solution for (2x2+6x+10)+(=4x2+2x+3) equation:

(2x^2+6x+10)+(=4x^2+2x+3)

We move all terms to the left:

(2x^2+6x+10)+(-(4x^2+2x+3))=0

We get rid of parentheses

2x^2+6x+(-(4x^2+2x+3))+10=0


We calculate terms in parentheses: +(-(4x^2+2x+3)), so:

-(4x^2+2x+3)

We get rid of parentheses

-4x^2-2x-3

Back to the equation:

+(-4x^2-2x-3)


We get rid of parentheses

2x^2-4x^2-2x+6x-3+10=0

We add all the numbers together, and all the variables

-2x^2+4x+7=0

a = -2; b = 4; c = +7;
Δ = b2-4ac
Δ = 42-4·(-2)·7
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6\sqrt{2}}{2*-2}=\frac{-4-6\sqrt{2}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6\sqrt{2}}{2*-2}=\frac{-4+6\sqrt{2}}{-4}
$