(2x-7)x+(6x-3)+(7x+10)=180

Solution for (2x-7)x+(6x-3)+(7x+10)=180 equation:

(2x-7)x+(6x-3)+(7x+10)=180

We move all terms to the left:

(2x-7)x+(6x-3)+(7x+10)-(180)=0

We multiply parentheses

2x^2-7x+(6x-3)+(7x+10)-180=0

We get rid of parentheses

2x^2-7x+6x+7x-3+10-180=0

We add all the numbers together, and all the variables

2x^2+6x-173=0

a = 2; b = 6; c = -173;
Δ = b2-4ac
Δ = 62-4·2·(-173)
Δ = 1420
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:

$\sqrt{\Delta}=\sqrt{1420}=\sqrt{4*355}=\sqrt{4}*\sqrt{355}=2\sqrt{355}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{355}}{2*2}=\frac{-6-2\sqrt{355}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{355}}{2*2}=\frac{-6+2\sqrt{355}}{4}
$