(5x+0.2)+(9-0.5x)+(6x-0.6)=(5x+1.1)(7x-0.3)+(8.5-2x)

Solution for (5x+0.2)+(9-0.5x)+(6x-0.6)=(5x+1.1)(7x-0.3)+(8.5-2x) equation:

(5x+0.2)+(9-0.5x)+(6x-0.6)=(5x+1.1)(7x-0.3)+(8.5-2x)

We move all terms to the left:

(5x+0.2)+(9-0.5x)+(6x-0.6)-((5x+1.1)(7x-0.3)+(8.5-2x))=0

We add all the numbers together, and all the variables

(5x+0.2)+(-0.5x+9)+(6x-0.6)-((5x+1.1)(7x-0.3)+(-2x+8.5))=0

We get rid of parentheses

5x-0.5x+6x-((5x+1.1)(7x-0.3)+(-2x+8.5))+0.2+9-0.6=0

We multiply parentheses ..

-((+35x^2-1.5x+7.7x-0.33)+(-2x+8.5))+5x-0.5x+6x+0.2+9-0.6=0


We calculate terms in parentheses: -((+35x^2-1.5x+7.7x-0.33)+(-2x+8.5)), so:

(+35x^2-1.5x+7.7x-0.33)+(-2x+8.5)

We get rid of parentheses

35x^2-1.5x+7.7x-2x-0.33+8.5

We add all the numbers together, and all the variables

35x^2+4.2x+8.17

Back to the equation:

-(35x^2+4.2x+8.17)


We add all the numbers together, and all the variables

10.5x-(35x^2+4.2x+8.17)+8.6=0

We get rid of parentheses

-35x^2+10.5x-4.2x-8.17+8.6=0

We add all the numbers together, and all the variables

-35x^2+6.3x+0.43=0

a = -35; b = 6.3; c = +0.43;
Δ = b2-4ac
Δ = 6.32-4·(-35)·0.43
Δ = 99.89
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{-(6.3)-\sqrt{99.89}}{2*-35}=\frac{-6.3-\sqrt{99.89}}{-70}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6.3)+\sqrt{99.89}}{2*-35}=\frac{-6.3+\sqrt{99.89}}{-70}
$