490^2=4x^2+2*(4x+3x)

Solution for 490^2=4x^2+2*(4x+3x) equation:

490^2=4x^2+2(4x+3x)

We move all terms to the left:

490^2-(4x^2+2(4x+3x))=0

We add all the numbers together, and all the variables

-(4x^2+2(+7x))+490^2=0

We add all the numbers together, and all the variables

-(4x^2+2(+7x))+240100=0


We calculate terms in parentheses: -(4x^2+2(+7x)), so:

4x^2+2(+7x)

We multiply parentheses

4x^2+14x

Back to the equation:

-(4x^2+14x)


We get rid of parentheses

-4x^2-14x+240100=0

a = -4; b = -14; c = +240100;
Δ = b2-4ac
Δ = -142-4·(-4)·240100
Δ = 3841796
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:

$\sqrt{\Delta}=\sqrt{3841796}=\sqrt{196*19601}=\sqrt{196}*\sqrt{19601}=14\sqrt{19601}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-14\sqrt{19601}}{2*-4}=\frac{14-14\sqrt{19601}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+14\sqrt{19601}}{2*-4}=\frac{14+14\sqrt{19601}}{-8}
$