7(5-x)=20+2(3+x)x=

Solution for 7(5-x)=20+2(3+x)x= equation:

7(5-x)=20+2(3+x)x=

We move all terms to the left:

7(5-x)-(20+2(3+x)x)=0

We add all the numbers together, and all the variables

7(-1x+5)-(20+2(x+3)x)=0

We multiply parentheses

-7x-(20+2(x+3)x)+35=0


We calculate terms in parentheses: -(20+2(x+3)x), so:

20+2(x+3)x

determiningTheFunctionDomain
2(x+3)x+20

We multiply parentheses

2x^2+6x+20

Back to the equation:

-(2x^2+6x+20)


We get rid of parentheses

-2x^2-7x-6x-20+35=0

We add all the numbers together, and all the variables

-2x^2-13x+15=0

a = -2; b = -13; c = +15;
Δ = b2-4ac
Δ = -132-4·(-2)·15
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-17}{2*-2}=\frac{-4}{-4}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+17}{2*-2}=\frac{30}{-4}
=-7+1/2
$