45(x+4)+45(x-4)=7(x-4)(x+4)

Solution for 45(x+4)+45(x-4)=7(x-4)(x+4) equation:

45(x+4)+45(x-4)=7(x-4)(x+4)

We move all terms to the left:

45(x+4)+45(x-4)-(7(x-4)(x+4))=0

We use the square of the difference formula

x^2+45(x+4)+45(x-4)+16=0

We multiply parentheses

x^2+45x+45x+180-180+16=0

We add all the numbers together, and all the variables

x^2+90x+16=0

a = 1; b = 90; c = +16;
Δ = b2-4ac
Δ = 902-4·1·16
Δ = 8036
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:

$\sqrt{\Delta}=\sqrt{8036}=\sqrt{196*41}=\sqrt{196}*\sqrt{41}=14\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(90)-14\sqrt{41}}{2*1}=\frac{-90-14\sqrt{41}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(90)+14\sqrt{41}}{2*1}=\frac{-90+14\sqrt{41}}{2}
$