(2d-5)(2d+5)=-10

Solution for (2d-5)(2d+5)=-10 equation:

(2d-5)(2d+5)=-10

We move all terms to the left:

(2d-5)(2d+5)-(-10)=0

We add all the numbers together, and all the variables

(2d-5)(2d+5)+10=0

We use the square of the difference formula

4d^2-25+10=0

We add all the numbers together, and all the variables

4d^2-15=0

a = 4; b = 0; c = -15;
Δ = b2-4ac
Δ = 02-4·4·(-15)
Δ = 240
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{15}}{2*4}=\frac{0-4\sqrt{15}}{8}
=-\frac{4\sqrt{15}}{8}
=-\frac{\sqrt{15}}{2}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{15}}{2*4}=\frac{0+4\sqrt{15}}{8}
=\frac{4\sqrt{15}}{8}
=\frac{\sqrt{15}}{2}
$