(7x-4)(x+4)+20=25

Solution for (7x-4)(x+4)+20=25 equation:

(7x-4)(x+4)+20=25

We move all terms to the left:

(7x-4)(x+4)+20-(25)=0

We add all the numbers together, and all the variables

(7x-4)(x+4)-5=0

We multiply parentheses ..

(+7x^2+28x-4x-16)-5=0

We get rid of parentheses

7x^2+28x-4x-16-5=0

We add all the numbers together, and all the variables

7x^2+24x-21=0

a = 7; b = 24; c = -21;
Δ = b2-4ac
Δ = 242-4·7·(-21)
Δ = 1164
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:

$\sqrt{\Delta}=\sqrt{1164}=\sqrt{4*291}=\sqrt{4}*\sqrt{291}=2\sqrt{291}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-2\sqrt{291}}{2*7}=\frac{-24-2\sqrt{291}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+2\sqrt{291}}{2*7}=\frac{-24+2\sqrt{291}}{14}
$