56=(3x)2+(5x-4)2

Solution for 56=(3x)2+(5x-4)2 equation:

56=(3x)2+(5x-4)2

We move all terms to the left:

56-((3x)2+(5x-4)2)=0


We calculate terms in parentheses: -(3x2+(5x-4)2), so:

3x2+(5x-4)2

We add all the numbers together, and all the variables

3x^2+(5x-4)2

We multiply parentheses

3x^2+10x-8

Back to the equation:

-(3x^2+10x-8)


We get rid of parentheses

-3x^2-10x+8+56=0

We add all the numbers together, and all the variables

-3x^2-10x+64=0

a = -3; b = -10; c = +64;
Δ = b2-4ac
Δ = -102-4·(-3)·64
Δ = 868
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:

$\sqrt{\Delta}=\sqrt{868}=\sqrt{4*217}=\sqrt{4}*\sqrt{217}=2\sqrt{217}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{217}}{2*-3}=\frac{10-2\sqrt{217}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{217}}{2*-3}=\frac{10+2\sqrt{217}}{-6}
$