45-(2x+3x+5x)x=4

Solution for 45-(2x+3x+5x)x=4 equation:

45-(2x+3x+5x)x=4

We move all terms to the left:

45-(2x+3x+5x)x-(4)=0

We add all the numbers together, and all the variables

-(+10x)x+45-4=0

We add all the numbers together, and all the variables

-(+10x)x+41=0

We multiply parentheses

-10x^2+41=0

a = -10; b = 0; c = +41;
Δ = b2-4ac
Δ = 02-4·(-10)·41
Δ = 1640
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:

$\sqrt{\Delta}=\sqrt{1640}=\sqrt{4*410}=\sqrt{4}*\sqrt{410}=2\sqrt{410}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{410}}{2*-10}=\frac{0-2\sqrt{410}}{-20}
=-\frac{2\sqrt{410}}{-20}
=-\frac{\sqrt{410}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{410}}{2*-10}=\frac{0+2\sqrt{410}}{-20}
=\frac{2\sqrt{410}}{-20}
=\frac{\sqrt{410}}{-10}
$