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

Solution for (x+5)(x-5)+7=0 equation:

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

We use the square of the difference formula

x^2-25+7=0

We add all the numbers together, and all the variables

x^2-18=0

a = 1; b = 0; c = -18;
Δ = b2-4ac
Δ = 02-4·1·(-18)
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{2}}{2*1}=\frac{0-6\sqrt{2}}{2}
=-\frac{6\sqrt{2}}{2}
=-3\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{2}}{2*1}=\frac{0+6\sqrt{2}}{2}
=\frac{6\sqrt{2}}{2}
=3\sqrt{2}
$