(7x2-4x-7)=(2x+8)

Solution for (7x2-4x-7)=(2x+8) equation:

(7x^2-4x-7)=(2x+8)

We move all terms to the left:

(7x^2-4x-7)-((2x+8))=0

We get rid of parentheses

7x^2-4x-((2x+8))-7=0


We calculate terms in parentheses: -((2x+8)), so:

(2x+8)

We get rid of parentheses

2x+8

Back to the equation:

-(2x+8)


We get rid of parentheses

7x^2-4x-2x-8-7=0

We add all the numbers together, and all the variables

7x^2-6x-15=0

a = 7; b = -6; c = -15;
Δ = b2-4ac
Δ = -62-4·7·(-15)
Δ = 456
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:

$\sqrt{\Delta}=\sqrt{456}=\sqrt{4*114}=\sqrt{4}*\sqrt{114}=2\sqrt{114}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{114}}{2*7}=\frac{6-2\sqrt{114}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{114}}{2*7}=\frac{6+2\sqrt{114}}{14}
$