-2x^2+71=(x+5)(x-5)

Solution for -2x^2+71=(x+5)(x-5) equation:

-2x^2+71=(x+5)(x-5)

We move all terms to the left:

-2x^2+71-((x+5)(x-5))=0

We use the square of the difference formula

-2x^2+x^2+25+71=0

We add all the numbers together, and all the variables

-1x^2+96=0

a = -1; b = 0; c = +96;
Δ = b2-4ac
Δ = 02-4·(-1)·96
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{6}}{2*-1}=\frac{0-8\sqrt{6}}{-2}
=-\frac{8\sqrt{6}}{-2}
=-\frac{4\sqrt{6}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{6}}{2*-1}=\frac{0+8\sqrt{6}}{-2}
=\frac{8\sqrt{6}}{-2}
=\frac{4\sqrt{6}}{-1}
$