40=(0.3x)x+4x

Solution for 40=(0.3x)x+4x equation:

40=(0.3x)x+4x

We move all terms to the left:

40-((0.3x)x+4x)=0

We add all the numbers together, and all the variables

-((+0.3x)x+4x)+40=0


We calculate terms in parentheses: -((+0.3x)x+4x), so:

(+0.3x)x+4x

We add all the numbers together, and all the variables

4x+(+0.3x)x

We multiply parentheses

0x^2+4x

We add all the numbers together, and all the variables

x^2+4x

Back to the equation:

-(x^2+4x)


We get rid of parentheses

-x^2-4x+40=0

We add all the numbers together, and all the variables

-1x^2-4x+40=0

a = -1; b = -4; c = +40;
Δ = b2-4ac
Δ = -42-4·(-1)·40
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{11}}{2*-1}=\frac{4-4\sqrt{11}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{11}}{2*-1}=\frac{4+4\sqrt{11}}{-2}
$