(x+3)+x=7x^2+3x/10

Solution for (x+3)+x=7x^2+3x/10 equation:

(x+3)+x=7x^2+3x/10

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-7x^2+x+x-3x/10+3=0

We multiply all the terms by the denominator


-7x^2*10+x*10+x*10-3x+3*10=0

We add all the numbers together, and all the variables

-7x^2*10-3x+x*10+x*10+30=0

Wy multiply elements

-70x^2-3x+10x+10x+30=0

We add all the numbers together, and all the variables

-70x^2+17x+30=0

a = -70; b = 17; c = +30;
Δ = b2-4ac
Δ = 172-4·(-70)·30
Δ = 8689
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{-(17)-\sqrt{8689}}{2*-70}=\frac{-17-\sqrt{8689}}{-140}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{8689}}{2*-70}=\frac{-17+\sqrt{8689}}{-140}
$