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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

5x^2+15x-(3(x-3)-4x)=0


We calculate terms in parentheses: -(3(x-3)-4x), so:

3(x-3)-4x

We add all the numbers together, and all the variables

-4x+3(x-3)

We multiply parentheses

-4x+3x-9

We add all the numbers together, and all the variables

-1x-9

Back to the equation:

-(-1x-9)


We get rid of parentheses

5x^2+15x+1x+9=0

We add all the numbers together, and all the variables

5x^2+16x+9=0

a = 5; b = 16; c = +9;
Δ = b2-4ac
Δ = 162-4·5·9
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{19}}{2*5}=\frac{-16-2\sqrt{19}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{19}}{2*5}=\frac{-16+2\sqrt{19}}{10}
$