40-(3x+4)=4(x+5)x

Solution for 40-(3x+4)=4(x+5)x equation:

40-(3x+4)=4(x+5)x

We move all terms to the left:

40-(3x+4)-(4(x+5)x)=0

We get rid of parentheses

-3x-(4(x+5)x)-4+40=0


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

4(x+5)x

We multiply parentheses

4x^2+20x

Back to the equation:

-(4x^2+20x)


We add all the numbers together, and all the variables

-3x-(4x^2+20x)+36=0

We get rid of parentheses

-4x^2-3x-20x+36=0

We add all the numbers together, and all the variables

-4x^2-23x+36=0

a = -4; b = -23; c = +36;
Δ = b2-4ac
Δ = -232-4·(-4)·36
Δ = 1105
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-\sqrt{1105}}{2*-4}=\frac{23-\sqrt{1105}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{1105}}{2*-4}=\frac{23+\sqrt{1105}}{-8}
$