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

Solution for (4x+1)(3x-7)-5(2-7x)=2x(x+10)-7 equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

(4x+1)(3x-7)+35x-(2x(x+10)-7)-10=0

We multiply parentheses ..

(+12x^2-28x+3x-7)+35x-(2x(x+10)-7)-10=0


We calculate terms in parentheses: -(2x(x+10)-7), so:

2x(x+10)-7

We multiply parentheses

2x^2+20x-7

Back to the equation:

-(2x^2+20x-7)


We get rid of parentheses

12x^2-2x^2-28x+3x+35x-20x-7+7-10=0

We add all the numbers together, and all the variables

10x^2-10x-10=0

a = 10; b = -10; c = -10;
Δ = b2-4ac
Δ = -102-4·10·(-10)
Δ = 500
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:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10\sqrt{5}}{2*10}=\frac{10-10\sqrt{5}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10\sqrt{5}}{2*10}=\frac{10+10\sqrt{5}}{20}
$