10x(4-x)=7x+2(3x+5)

Solution for 10x(4-x)=7x+2(3x+5) equation:

10x(4-x)=7x+2(3x+5)

We move all terms to the left:

10x(4-x)-(7x+2(3x+5))=0

We add all the numbers together, and all the variables

10x(-1x+4)-(7x+2(3x+5))=0

We multiply parentheses

-10x^2+40x-(7x+2(3x+5))=0


We calculate terms in parentheses: -(7x+2(3x+5)), so:

7x+2(3x+5)

We multiply parentheses

7x+6x+10

We add all the numbers together, and all the variables

13x+10

Back to the equation:

-(13x+10)


We get rid of parentheses

-10x^2+40x-13x-10=0

We add all the numbers together, and all the variables

-10x^2+27x-10=0

a = -10; b = 27; c = -10;
Δ = b2-4ac
Δ = 272-4·(-10)·(-10)
Δ = 329
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{-(27)-\sqrt{329}}{2*-10}=\frac{-27-\sqrt{329}}{-20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+\sqrt{329}}{2*-10}=\frac{-27+\sqrt{329}}{-20}
$